Sequences for Calculus 2 (Section 11.1)

welcome to section 11 point 1 on sequences for calculus 2 so first we're gonna begin with the definition a sequence is a function whose domain is the set of natural numbers and the way we usually express a sequence or representing is by writing a sub N and notice carefully that the definition says the domain of a sequence is the set of natural numbers so natural numbers are the numbers 1 2 3 all of the positive integers there's a little bit about of debate about 0 above anyways you can think of it as a of n being a function right of n so all the values that get substituted in as part of the domain have to be natural numbers we're not allowing all real numbers rational numbers or anything like that now a sequence is a list of numbers a sub 1 a sub 2 a sub 3 so you represent which term in the sequence you're referring to by the subscript and that also corresponds with the natural number that gets substituted in for n into the expression for the sequence and the graph consists of isolated points whose coordinates would be 1 a sub 1 2 a sub 2 etc it can be finite or infinite and we're just gonna graph some arbitrary sequence for you to get a little idea so here this would be my n access and this I'm gonna label a sub n those are all the sequence values and I'll start maybe will list about 5 points and notice this graph all of the points are gonna be discrete so I have a sub 1 there's just a single point somewhere a sub 2 may be a sub 3 will put down there a sub 4 a sub byte so you get the idea this is how we would represent the graph of a sequence now let's move on to some examples of specific sequences and how we work with them so first example find a formula for a sub n a sub n is called the general term alright so here they've listed out the first few terms of this sequence so this would be a sub one right our first term this is a sub 2 a sub 3 and hopefully as you look at the different terms you can see a pattern for what's going on this is actually a geometric sequence you may have heard that before from your previous courses but the most important thing to notice is that to get from one term to the next you're multiplying by 1/3 every single time so that's your common ratio well how would I give a formula so that someone could come up with the hundredth term or some general term well I want to take 1/3 right and raise it to some power but I have to be careful because I want the first term to come out to be 1 and I want the second term to come out to be 1/3 so I'm gonna take 1/3 and raise it to the N minus 1 that way it'll match all of the terms that were given for this sequence exactly all right and you can always check to if you're not certain so we can check here if I want to substitute in 1 for n then we'll have 1/3 to the 1 minus 1 so that's 1/3 to the 0 so that's one and then a sub let's skip around a sub 3 that would be 1/3 to the 3 minus 1 so 1/3 squared and that's one night so everything checks out all right let's look at another example again the directions are the same so find a formula for a sub n the general term and this time the sequence looks a little different the first term is 5 then 1 then 5 then 1 and so on and so forth you can just expect the pattern to stay consistent so this this sequence is oscillating right between two numbers 5 & 1 and there's more than one way to represent this but probably the cleanest way is by first identifying what the halfway point is between these two numbers so halfway between 5 and one is positive three so you could think of that as the midline and then the sequence either takes three and adds two or it takes three and subtracts 2 that's how you get from one term to the next all right now how can i construct a formula in order to express this well it looks like on the first the third the fifth all of the odd terms I want to add to and then on all of the even terms right the second term the fourth term the sixth term I want to subtract two so I need to either add to or subtract two and I need the oscillation to occur based on whether I have an odd or an even term so the way I do that is I'm gonna use negative one raised to some power and remember the baseline is free that middle line so I have three plus and I want the first term to end up adding two so I'm gonna have negative one raised to the N minus one times two and then this is gonna be our formula for the general term so let's go through and break it down piece by piece we start off with three and then if I have negative 1 to the N minus one then all of the odd number terms negative one's gonna be raised to an even power so it's gonna be positive times two so I'm gonna end up adding two and then on all of the even number terms and minus one is going to be odd so I'm going to have negative one times two so it's going to end up subtracting two and we can just double check and confirm by substituting in a few values for n all right so let's make sure for the first term a sub one that's going to be 3 plus negative 1 to the 1 minus 1 times 2 so that's gonna be negative 1 to the 0 which is 1 so this is 3 plus 2 which is 5 okay looks great now let's make sure the second term still works so if a sub 2 we're gonna have 3 plus negative 1 to the 2 minus 1 times 2 so negative 1 to the 2 minus 1 that's going to be negative 1 times 2 so this is going to be 3 minus - which is one and the pattern will continue so that one is complete now just to extend the problem what if we had almost the exact same sequence but noticed I'm gonna change something up 1/5 1/5 etc ok so what I would want now is for the first term the third term the fifth term all the odd number terms for two to be subtracted and for all the even number terms the second the fourth to have two added well I would just need that negative one to be raised to a different power so what I would do is raise the negative 1 to the nth power and then times 2 and then that would switch so that all of the odd number terms would end up subtracting - and the even number terms would add - all right good moving on some more definitions all right so we say the sequence a sub n has the limit L if the limit as n approaches infinity of a sub n is equal to L and a convergent sequence is one that satisfies the condition that the limit as n approaches infinity of a sub n exists and that limit have to exist as a finite number a divergent sequence is one where the limit as n approaches infinity of a sub n either does not exist or it equals positive or negative infinity so convergent sequences are only sequences whose infinite limit equals a finite number now we have some limit laws for convergent sequences and you might notice that they're very similar to our limit laws for functions the thing to remember though is that these limit laws can only be applied to convergent sequences so we're only allowed to use them if we know that these sequences a sub N and B sub n are convergent so we're allowed to distribute a limit over addition or subtraction as well as multiplication and division the denominator is not approaching zero the fourth limit logs very powerful we're allowed to take a constant times a sequence and bring it outside of the limit the limit as n approaches infinity of a constant it's just a constant and I can also raise it to a power all right next theorem if the limit of n approaches infinity of a sub n equals L and s is continuous at L then the limit as n approaches infinity of f of a sub n is equal to F of L so what this means is if I have a continuous function then I can pass the limit through the composition of that function with our sequence all right and this is going to be very useful as we try to evaluate limits of more complicated or involved sequences so here's an example to demonstrate example 3 determine whether the sequence converges or diverges and then if it converges find out all right so to begin I'm just gonna take the limit as n approaches infinity of a sub n so this is the limit as n approaches infinity of the nth root of 2 raised to the 1 + 3 n all right looks good instead of writing the nth root I can rewrite this as the limit as n approaches infinity of 2 raised to the 1 + 3 N and then that quantity is raised to the 1 over N and then using my laws of exponents I can rewrite this now as the limit as n approaches infinity of 2 when you have a power raised to another power you multiply the exponents so this is gonna be 1 + 3 n times 1 over N or divided by head all right well I'm noticing here - my base is a constant right so in order to evaluate this limit I want to see where the exponent is approaching so I want to pass the limit through + 2 raised to some power that's an exponential function that's continuous as n approaches positive infinity so I'm allowed to pass this limit through to the exponent thanks to the theorem that we just considered so this is equal to 2 raised to the limit as n approaches infinity of 1 plus 3n over N and then remembering when we take infinite limits since the highest power in the numerator is equal to the highest power in the denominator then I can just divide the coefficients of our leading terms and so the exponent is going to be approaching 3 which means now this all simplifies to 2 to the 3rd which is 8 and this is actually what L is or our limit is equal to all right and now make sure you answer the question completely and asked to determine whether the sequence converges or diverges so we can say now since the limit exists as a finite number the sequence converges both parts are required for a complete answer alright that's looking good let's move on another theorem if the limit as n approaches infinity of the absolute value of f of n equals 0 then the limit of n approaches infinity of a sub N equals 0 this theorem is super powerful and useful so make sure you remember the statement and we're going to look at an example now on how to apply it so example for the limit as n approaches infinity of negative 1 to the N over N now if I try to take this limit directly negative 1 to the N oscillates right it's gonna isolate between positive 1 and negative 1 so I can't evaluate this limit because of that alternating term however what I'm gonna do is consider the limit as n approaches infinity of the absolute value of negative 1 to the N over N and when I take the absolute value of negative 1 to the N that's just going to end up giving me positive 1 so this is the same as the limit as n approaches infinity of just 1 over n on this limit I can easily evaluate as n goes to infinity the denominator is getting larger and larger so the terms approach zero and based on this theorem since I know that the absolute value of my original sequence that limit equals zero then I know the limit as n approaches infinity of negative 1 to the N over N must also equal 0 as well all right let's look at a graph of what this sequence looks like to help reinforce this idea okay so here's the N axis it's a little crooked let's make it perfect okay here's the N axis here's a sub n here's one here's negative 1 so here's 8 my first term is gonna go here second term third term all right so a sub 1 remember we're gonna substitute everything into a sub N equals negative 1 to the N over N so a sub 1 is negative 1 it's gonna be down here a sub 2 if I had negative 1 squared that's positive 1 1 over 2 that's 1/2 so that'll be here a sub 3 that's gonna be negative 1/3 a sub 4 positive 1/4 and negative 1/5 etc etc so we see it's oscillating for sure right it alternates between being positive and negative but since the terms are shrinking the limit is still going to be going to 0 and I can see that because even if I were to take the absolute value of all the terms so if I took the absolute value of a sub 1 and it was up here and if I took the absolute value of a sub 3 and it was up here we could still see that the terms are shrinking and approaching 0 all right good moving on the squeeze theorem for sequences this is going to be very similar to the squeeze theorem that you use for functions so if a sub n is less than or equal to B sub n is less than or equal to C sub n for all N greater than or equal to n naught and if the limit as n approaches infinity of a sub N and the limit as n approaches infinity of C sub n both equal al then the limit of as n approaches infinity of B sub n also equals out so here's our sequence whose limit we're trying to evaluate and if I can bound it by two other sequences who have the same limit then I know that the limit of our sequence must also be the same this is great to use for functions that can or sequences that can be easily bounded so with the easiest kind of sequence to bound something involving sine and cosine usually usually so here's an example we're gonna figure out if the sequence is convergent or divergent and we have a sub n which is sine of 2 n over 1 plus rad end now if I were to just try to take the limit as n approaches infinity directly I wouldn't be really successful because sine of 2n oscillates remember the graph of sine is a wave so I couldn't proceed from there but what I'm gonna do is use the squeeze theorem so I can bound sine of 2n between 1 and negative 1 right remember your graph of sine its amplitude is 1 so all values are bounded between 1 and negative 1 now what I want to try to do is get this quantity in the middle to match my sequence whose limit I'm trying to find alright so what I need to do is divide through by 1 plus radda now you have to be careful when you're working with an inequality rad n is always going to be positive and if I add 1 that's going to be positive 2 so I don't need to worry about flipping any directions of inequality signs but consider it okay so we have negative 1 over 1 plus R at n is less than or equal to sine of 2 n over 1 plus R at n which is less than or equal to 1 over 1 Ravin all right so here's our sequence whose limit we're trying to evaluate and this is basically a sub n from the squeeze theorem and C sub n so I want to take the limit of each of those sequences as n approaches infinity and as long as I get the same value then that also the limit of our sequence so let's see the limit as n approaches infinity of negative 1 over 1 plus R at n well the denominator is just going to get larger and larger numerators are constant so this is 0 and the limit as n approaches infinity of 1 over 1 plus R at n we'll same scenario right the numerator is a constant denominators getting very very large this is also 0 so what does this tell me this implies that the limit as n approaches infinity of sine 2n over 1 plus R at n also equals 0 and you must state by the squeeze theorem all right so it's very important to remember anytime you apply the squeeze theorem you have to rewrite the original sequence with proper limit notation and state that you used the squeeze theorem all right looks nice moving on to the next idea theorem if the limit as X approaches infinity of f of X so now we're talking about a real valued function equals L and f of N equals a sub n so that's our sequence where N is a natural number then the limit as n approaches infinity of f of n is equal to the limit as n approaches infinity of a sub n which is out so what this basically tells us is if we have a function whose infinite limit is L and that function can also be used to represent our sequence but just remember now our domain is going to be the set of natural numbers then they would have the same infinite limit and this is going to be really useful so that we can apply additional limit techniques so let's consider an example here a sub n is equal to natural log of N squared over N so just at first glance when you're taking a limit for sequences we always take the limit to infinity and I noticed that the numerator is approaching infinity and the denominator is also approaching infinity as n gets very very large so I have an indeterminate form of the type infinity over infinity and one of our first inclination is to use l'hopital's rule but remember we're only allowed to use l'hopital's rule as long as both the numerator and denominator of our expression are differentiable functions and we know that sequences are definitely not differentiable the domain is a discrete set it's a set of natural numbers so what I have to do is define a function I'm gonna say let s of X equal and then I'm just gonna copy exactly the same sequence but replace all the ends with X's so then we're gonna have the natural log of x squared over X now I'm going to consider the limit as X approaches infinity of this function and I know from my theorem before whatever the limit of this function is it's going to be the same as the limit of our sequence so this is the limit as X approaches infinity of the natural log of x squared over X again as X gets arbitrarily large the numerators approaching infinity denominators also approaching infinity so I'm going to apply l'hopital's rule make sure you indicate that you're doing so so this is gonna be the limit as X approaches infinity derivative of the numerator so I'm gonna use the power rule I have two times the natural log of X and then by the chain rule I have to multiply by the derivative natural log of X which is 1 over x over and then the derivative of the denominator the derivative of X is just 1 all right so I can rewrite this as the limit as X approaches infinity of 2 Ln of X over X all right now let's investigate what's going on so as X approaches infinity the numerator is approaching infinity the denominator is also approaching infinity so again I have an indeterminate form of the type infinity over infinity so I can apply l'hopital's rule again and we have the limit as X approaches infinity derivative of 2 times the natural log of X is just going to be 2 over x over derivative of X is just 1 so now I can rewrite all of this as the limit as X approaches infinity of 2 over X and if I have a constant over the denominators approaching infinity this is equal to 0 so what is this all Tommy well basically now I can say all right based on our theorem the limit as n approaches infinity of a sub n is equal to the limit as n approaches infinity of the natural log of N squared over n which also equals 0 so we would say this sequence converges all right make sure anytime you do this anytime you apply l'hopital's rule you first define a function f of X otherwise your work is incorrect because l'hopital's rule is not valid unless you have a differentiable function so you cannot apply it to a sequence directly all right moving on another fact say we have the limit as n approaches infinity of R to the N and R is between negative 1 and 1 then that limit is going to be 0 that should make sense because basically you're to get from 1 turn to another you're multiplying by something smaller than one so the sequence is going to be shrinking shrinking shrinking from one term to the next so it's going to get to zero eventually notice when art is equal to positive one the limit as n approaches infinity of R to the N would just be positive one and that converges also so here we have an example where we're going to try to apply this we have a sub n our sequence equals 3 to the n plus 2 divided by 5/2 then so let's see if I can rewrite it so I can identify what are that common ratio is and I'm gonna use my laws of exponents and break up the n plus 2 in the numerator and we write it as 3 to the N times 3 squared over 5 to the N and now I can group 3 to the N over 5 to the N and 3 squared that's just 9 right this is 9 so rewriting my sequence I have 9 times 3/5 raised to them so if you wanted to list off the first few terms to kind of get a feel for what's going on you would have 9 times 3/5 to the first and then 9 times 3/5 squared so that would be 9 over 25 and then 9 times 3/5 cubed which is 27 over 125 etc and you could see from one term to the next they're always going to be shrinking so the sequence is approaching zero about how do we justify that well we're going to take the limit as n approaches infinity of 3 to the n plus 2 over 5 to the M and we know now we can rewrite that as the limit as n approaches infinity of 9 times 3/5 to that and then using one of our limit laws I can take the 9 out side of the limit so I have 9 times the limit as n approaches infinity of 3/5 to the N since this our value is less than one then I know that that limit goes to zero and you want to indicate that this goes to zero and here's our justification make sure you include that so if that goes to zero then we have 9 times zero so this limit is zero and our sequence converges all right nice job couple more ideas let's summarize before we add on so first summary if you see that you have a geometric sequence you have some common ratio raised to a power you want to check if R is between negative 1 and 1 inclusive that'll let you know if it's convergent or not squeeze theorem that's especially useful for sine or cosine in a sequence if you have a sub n divided by B sub n where they're both polynomials numerator and denominator you can use l'hopital's rule if you change everything to be a function of X or you can just divide by the highest degree of the denominator now new definitions an increasing sequence is one where the first term a sub 1 is less than a sub 2 is less than a sub 3 for all N greater than or equal to 1 increasing if the terms get larger decreasing sequence if a sub 1 is greater than a sub 2 is greater than a sub 3 for all and greater than or equal to 1 a monotonic sequence is one that's either increasing or decreasing meaning that all the terms are either going to be increasing for the entire domain or always decreasing one or the other now we say a sequence is bounded above if there's some number H such that all of the terms are less than H for all N greater than or equal to one and the sequence is bounded below if there's a number lowercase M such that all a sub ends are greater than or equal to M for all n greater than or equal to one so basically all the terms are bigger than some number a bounded sequence is a sequence that has a upper bound H and the lower bound M so it's bounded above and below and the most exciting part of all of this is our theorem here that says every bounded monotonic sequence converges and I think the best way for you to really get a feel for that theorem is to look at a little graph so we're gonna graph some arbitrary sequence but remember it's bounded right so when we say a sequence is bounded that means it has both an upper bound and a lower bound it doesn't matter what they are so we'll say here's H our upper bound all the terms have to be less than H or equal to it and then our lower bound lower case and I'll just put it here okay so our sequence a sub n I don't know what it's doing but it lives between H and M okay it's only allowed to live in there and then the other part of the theorem says this sequence is also monotonic meaning it's either strictly increasing or strictly decreasing so pick a point to start it doesn't matter where I'm gonna call that one a sub one say it's strictly decreasing that means I'm only allowed when I move to the next term to decrease well there's no way for this sequence to not converge or to not have a limit of n approaches infinity if it's bounded it can't do something crazy like spiral down towards negative infinity because it's bounded below by lowercase M what if you started off somewhere house maybe down here and the sequence was increasing well again if it's strictly increasing it since it has to be monotonic it doesn't have the option of going back down it can only keep getting larger and larger so the limit would be capital H but notice you need a sequence to be bounded both above and below and be monotonic in order to apply this theorem alright so let's see how we can apply it consider the following sequence a sub n which equals 1 over n the terms are as follows if you want to list out a few to get a feel for it 1 1/2 1/3 1/4 now a sub n is bounded if I'm gonna make a statement like that I need to find what the upper bound and lower bound are well it's bounded above by 1 because that's the largest term that the sequence can ever be and it's bounded below by 0 right there's no way for me to get a negative term for a sub n since I'm only substituting in the natural numbers for n also this sequence is monotonic because it's strictly decreasing this means that it converges based on our theorem in fact we know that the limit as n approaches infinity of a sub n is equal to 0 now what if I didn't have all three of those conditions say I considered the following sequence a sub n which equals negative 1 to the n plus 1 well let's list out the first few terms so if I substitute in 1 for n that's gonna be negative 1 to the 1 plus 1 so negative 1 squared which is 1 then the next term would be negative 1 then 1 then negative 1 etc if a sub n bounded can you find an upper bound in the lower bound absolutely yes it's bounded it's bounded above by 1 right it's never gonna be larger than 1 and it's bounded below by negative 1 however is it monotonic wanted negative 1 1 negative 1 it switches from increasing to decreasing which means a sub n is not monotonic which means that a sub n does not converge in fact it diverges and if you want to see this this one's pretty fun one to graph because it's just gonna be bouncing between 1 and negative 1 right so here's n here's a sub n here's a positive 1 and negative 1 and all its gonna do is just bounce between those two so it'll be a positive 1 for the first term 1 2 3 4 5 then negative 1 then positive 1 then negative 1 and positive 1 etc so as n approaches infinity you could see the sequence isn't leveling off its not approaching any finite number alright last example here determine whether the sequence is increasing or decreasing or not monotonic is it banded alright so here we have a sequence a sub n which equals to n minus 3 over 3 n plus 4 well last time we had a really easy sequence just 1 over n we listed up the first few terms and we could clearly see that it was decreasing let's try listing out the first few terms here and see what happens so a sub 1 if I substitute in 1 for n that's gonna be 2 minus 3 over 3 plus 4 so that's negative 1 7 and then a sub 2 that would be 4 minus 3 over 6 plus 4 so that's one-tenth a sub 3 is 6 minus 3 over 9 plus 4 so that's gonna be 3 over 13 a sub 4 it's gonna give us 5 over 16 so it's not super obvious what's going on here what's another way that we could test whether something's increasing or decreasing well from calc 1 we know that the first derivative gives us information about where a function is increasing or decreasing but again I can't take a derivative of a sequence it's not differentiable so similarly to when we wanted to apply l'hopital's rule i'm gonna consider a function f of X and I'm gonna let it equal to X minus 3 over 3x plus 4 alright so redefine your sequence now of the function this is going to be differentiable so f prime of X is equal to so that's a need to use the quotient rule I'm gonna have 3x plus 4 times derivative of the numerator that's gonna be 2 minus 2x minus 3 times derivative of the denominators 3 and then square the denominator alright so cleaning up the numerator we're gonna have 6x Plus 8 minus if I distribute 3 that's gonna give me a 6x and then a positive 9 over 3x plus 4 squared and this is it being a really nice derivative because 6x cancels out so I just have 17 over 3x plus 4 squared well what can I say about this derivative the denominators squared so it's not going to be negative and the numerator is a positive constant so this is greater than 0 for all real numbers well what does this mean about our sequence if the derivative is positive that means my function is increasing strictly increasing since it's never equal to 0 so my sequence a sub n 2n minus 3 over 3 n plus 4 is increasing you could even say strictly increasing meaning it never is constant all right now let's go back to the problem and see what else it asks so we determine that the sequence is increasing is it bounded in order to answer that question I have to find the upper and lower bound if I'm unable to do so than I need to show that the sequence either approaches infinity or negative infinity by taking a limit so let's see the sequence is increasing then that means automatically I have the lower bound so the lower bound is gonna be the first term because the sequence keeps getting larger as you move to the next term so lower bound I got it it's just gonna be a sub 1 which is negative 1 7 because it's increasing great what about an upper bound can I find one well let's consider the limit as n approaches infinity of the sequence of 2 and minus 3 over 3 n plus 4 well here as n approaches infinity I look at the numerator and the denominator this is first degree in the numerator 1st degree in the denominator so I can just divide the leading coefficient and this limit is gonna be 2/3 so that means the sequence is approaching 2/3 as n goes to infinity it will never achieve that value of 2/3 so that's our upper bound so since I was able to find both an upper bound and a lower bound that means my sequence is indeed bounded alright so be sure anytime you use the derivative to check if a sequence is increasing or decreasing you redefine it as a function and in order to prove or state that it's bounded you need to find both the upper and lower bounds that concludes section 11 one stay tuned for 11 too

welcome to section 11 point 1 on sequences for calculus 2 so first we&#39;re gonna begin with the definition a sequence is a function whose domain is the set of natural numbers and the way we usually express a sequence or representing is by writing a sub N and notice carefully that the definition says the domain of a sequence is the set of natural numbers so natural numbers are the numbers 1 2 3 all of the positive integers there&#39;s a little bit about of debate about 0 above anyways you can think of it as a of n being a function right of n so all the values that get substituted in as part of the domain have to be natural numbers we&#39;re not allowing all real numbers rational numbers or anything like that now a sequence is a list of numbers a sub 1 a sub 2 a sub 3 so you represent which term in the sequence you&#39;re referring to by the subscript and that also corresponds with the natural number that gets substituted in for n into the expression for the sequence and the graph consists of isolated points whose coordinates would be 1 a sub 1 2 a sub 2 etc it can be finite or infinite and we&#39;re just gonna graph some arbitrary sequence for you to get a little idea so here this would be my n access and this I&#39;m gonna label a sub n those are all the sequence values and I&#39;ll start maybe will list about 5 points and notice this graph all of the points are gonna be discrete so I have a sub 1 there&#39;s just a single point somewhere a sub 2 may be a sub 3 will put down there a sub 4 a sub byte so you get the idea this is how we would represent the graph of a sequence now let&#39;s move on to some examples of specific sequences and how we work with them so first example find a formula for a sub n a sub n is called the general term alright so here they&#39;ve listed out the first few terms of this sequence so this would be a sub one right our first term this is a sub 2 a sub 3 and hopefully as you look at the different terms you can see a pattern for what&#39;s going on this is actually a geometric sequence you may have heard that before from your previous courses but the most important thing to notice is that to get from one term to the next you&#39;re multiplying by 1/3 every single time so that&#39;s your common ratio well how would I give a formula so that someone could come up with the hundredth term or some general term well I want to take 1/3 right and raise it to some power but I have to be careful because I want the first term to come out to be 1 and I want the second term to come out to be 1/3 so I&#39;m gonna take 1/3 and raise it to the N minus 1 that way it&#39;ll match all of the terms that were given for this sequence exactly all right and you can always check to if you&#39;re not certain so we can check here if I want to substitute in 1 for n then we&#39;ll have 1/3 to the 1 minus 1 so that&#39;s 1/3 to the 0 so that&#39;s one and then a sub let&#39;s skip around a sub 3 that would be 1/3 to the 3 minus 1 so 1/3 squared and that&#39;s one night so everything checks out all right let&#39;s look at another example again the directions are the same so find a formula for a sub n the general term and this time the sequence looks a little different the first term is 5 then 1 then 5 then 1 and so on and so forth you can just expect the pattern to stay consistent so this this sequence is oscillating right between two numbers 5 &amp; 1 and there&#39;s more than one way to represent this but probably the cleanest way is by first identifying what the halfway point is between these two numbers so halfway between 5 and one is positive three so you could think of that as the midline and then the sequence either takes three and adds two or it takes three and subtracts 2 that&#39;s how you get from one term to the next all right now how can i construct a formula in order to express this well it looks like on the first the third the fifth all of the odd terms I want to add to and then on all of the even terms right the second term the fourth term the sixth term I want to subtract two so I need to either add to or subtract two and I need the oscillation to occur based on whether I have an odd or an even term so the way I do that is I&#39;m gonna use negative one raised to some power and remember the baseline is free that middle line so I have three plus and I want the first term to end up adding two so I&#39;m gonna have negative one raised to the N minus one times two and then this is gonna be our formula for the general term so let&#39;s go through and break it down piece by piece we start off with three and then if I have negative 1 to the N minus one then all of the odd number terms negative one&#39;s gonna be raised to an even power so it&#39;s gonna be positive times two so I&#39;m gonna end up adding two and then on all of the even number terms and minus one is going to be odd so I&#39;m going to have negative one times two so it&#39;s going to end up subtracting two and we can just double check and confirm by substituting in a few values for n all right so let&#39;s make sure for the first term a sub one that&#39;s going to be 3 plus negative 1 to the 1 minus 1 times 2 so that&#39;s gonna be negative 1 to the 0 which is 1 so this is 3 plus 2 which is 5 okay looks great now let&#39;s make sure the second term still works so if a sub 2 we&#39;re gonna have 3 plus negative 1 to the 2 minus 1 times 2 so negative 1 to the 2 minus 1 that&#39;s going to be negative 1 times 2 so this is going to be 3 minus - which is one and the pattern will continue so that one is complete now just to extend the problem what if we had almost the exact same sequence but noticed I&#39;m gonna change something up 1/5 1/5 etc ok so what I would want now is for the first term the third term the fifth term all the odd number terms for two to be subtracted and for all the even number terms the second the fourth to have two added well I would just need that negative one to be raised to a different power so what I would do is raise the negative 1 to the nth power and then times 2 and then that would switch so that all of the odd number terms would end up subtracting - and the even number terms would add - all right good moving on some more definitions all right so we say the sequence a sub n has the limit L if the limit as n approaches infinity of a sub n is equal to L and a convergent sequence is one that satisfies the condition that the limit as n approaches infinity of a sub n exists and that limit have to exist as a finite number a divergent sequence is one where the limit as n approaches infinity of a sub n either does not exist or it equals positive or negative infinity so convergent sequences are only sequences whose infinite limit equals a finite number now we have some limit laws for convergent sequences and you might notice that they&#39;re very similar to our limit laws for functions the thing to remember though is that these limit laws can only be applied to convergent sequences so we&#39;re only allowed to use them if we know that these sequences a sub N and B sub n are convergent so we&#39;re allowed to distribute a limit over addition or subtraction as well as multiplication and division the denominator is not approaching zero the fourth limit logs very powerful we&#39;re allowed to take a constant times a sequence and bring it outside of the limit the limit as n approaches infinity of a constant it&#39;s just a constant and I can also raise it to a power all right next theorem if the limit of n approaches infinity of a sub n equals L and s is continuous at L then the limit as n approaches infinity of f of a sub n is equal to F of L so what this means is if I have a continuous function then I can pass the limit through the composition of that function with our sequence all right and this is going to be very useful as we try to evaluate limits of more complicated or involved sequences so here&#39;s an example to demonstrate example 3 determine whether the sequence converges or diverges and then if it converges find out all right so to begin I&#39;m just gonna take the limit as n approaches infinity of a sub n so this is the limit as n approaches infinity of the nth root of 2 raised to the 1 + 3 n all right looks good instead of writing the nth root I can rewrite this as the limit as n approaches infinity of 2 raised to the 1 + 3 N and then that quantity is raised to the 1 over N and then using my laws of exponents I can rewrite this now as the limit as n approaches infinity of 2 when you have a power raised to another power you multiply the exponents so this is gonna be 1 + 3 n times 1 over N or divided by head all right well I&#39;m noticing here - my base is a constant right so in order to evaluate this limit I want to see where the exponent is approaching so I want to pass the limit through + 2 raised to some power that&#39;s an exponential function that&#39;s continuous as n approaches positive infinity so I&#39;m allowed to pass this limit through to the exponent thanks to the theorem that we just considered so this is equal to 2 raised to the limit as n approaches infinity of 1 plus 3n over N and then remembering when we take infinite limits since the highest power in the numerator is equal to the highest power in the denominator then I can just divide the coefficients of our leading terms and so the exponent is going to be approaching 3 which means now this all simplifies to 2 to the 3rd which is 8 and this is actually what L is or our limit is equal to all right and now make sure you answer the question completely and asked to determine whether the sequence converges or diverges so we can say now since the limit exists as a finite number the sequence converges both parts are required for a complete answer alright that&#39;s looking good let&#39;s move on another theorem if the limit as n approaches infinity of the absolute value of f of n equals 0 then the limit of n approaches infinity of a sub N equals 0 this theorem is super powerful and useful so make sure you remember the statement and we&#39;re going to look at an example now on how to apply it so example for the limit as n approaches infinity of negative 1 to the N over N now if I try to take this limit directly negative 1 to the N oscillates right it&#39;s gonna isolate between positive 1 and negative 1 so I can&#39;t evaluate this limit because of that alternating term however what I&#39;m gonna do is consider the limit as n approaches infinity of the absolute value of negative 1 to the N over N and when I take the absolute value of negative 1 to the N that&#39;s just going to end up giving me positive 1 so this is the same as the limit as n approaches infinity of just 1 over n on this limit I can easily evaluate as n goes to infinity the denominator is getting larger and larger so the terms approach zero and based on this theorem since I know that the absolute value of my original sequence that limit equals zero then I know the limit as n approaches infinity of negative 1 to the N over N must also equal 0 as well all right let&#39;s look at a graph of what this sequence looks like to help reinforce this idea okay so here&#39;s the N axis it&#39;s a little crooked let&#39;s make it perfect okay here&#39;s the N axis here&#39;s a sub n here&#39;s one here&#39;s negative 1 so here&#39;s 8 my first term is gonna go here second term third term all right so a sub 1 remember we&#39;re gonna substitute everything into a sub N equals negative 1 to the N over N so a sub 1 is negative 1 it&#39;s gonna be down here a sub 2 if I had negative 1 squared that&#39;s positive 1 1 over 2 that&#39;s 1/2 so that&#39;ll be here a sub 3 that&#39;s gonna be negative 1/3 a sub 4 positive 1/4 and negative 1/5 etc etc so we see it&#39;s oscillating for sure right it alternates between being positive and negative but since the terms are shrinking the limit is still going to be going to 0 and I can see that because even if I were to take the absolute value of all the terms so if I took the absolute value of a sub 1 and it was up here and if I took the absolute value of a sub 3 and it was up here we could still see that the terms are shrinking and approaching 0 all right good moving on the squeeze theorem for sequences this is going to be very similar to the squeeze theorem that you use for functions so if a sub n is less than or equal to B sub n is less than or equal to C sub n for all N greater than or equal to n naught and if the limit as n approaches infinity of a sub N and the limit as n approaches infinity of C sub n both equal al then the limit of as n approaches infinity of B sub n also equals out so here&#39;s our sequence whose limit we&#39;re trying to evaluate and if I can bound it by two other sequences who have the same limit then I know that the limit of our sequence must also be the same this is great to use for functions that can or sequences that can be easily bounded so with the easiest kind of sequence to bound something involving sine and cosine usually usually so here&#39;s an example we&#39;re gonna figure out if the sequence is convergent or divergent and we have a sub n which is sine of 2 n over 1 plus rad end now if I were to just try to take the limit as n approaches infinity directly I wouldn&#39;t be really successful because sine of 2n oscillates remember the graph of sine is a wave so I couldn&#39;t proceed from there but what I&#39;m gonna do is use the squeeze theorem so I can bound sine of 2n between 1 and negative 1 right remember your graph of sine its amplitude is 1 so all values are bounded between 1 and negative 1 now what I want to try to do is get this quantity in the middle to match my sequence whose limit I&#39;m trying to find alright so what I need to do is divide through by 1 plus radda now you have to be careful when you&#39;re working with an inequality rad n is always going to be positive and if I add 1 that&#39;s going to be positive 2 so I don&#39;t need to worry about flipping any directions of inequality signs but consider it okay so we have negative 1 over 1 plus R at n is less than or equal to sine of 2 n over 1 plus R at n which is less than or equal to 1 over 1 Ravin all right so here&#39;s our sequence whose limit we&#39;re trying to evaluate and this is basically a sub n from the squeeze theorem and C sub n so I want to take the limit of each of those sequences as n approaches infinity and as long as I get the same value then that also the limit of our sequence so let&#39;s see the limit as n approaches infinity of negative 1 over 1 plus R at n well the denominator is just going to get larger and larger numerators are constant so this is 0 and the limit as n approaches infinity of 1 over 1 plus R at n we&#39;ll same scenario right the numerator is a constant denominators getting very very large this is also 0 so what does this tell me this implies that the limit as n approaches infinity of sine 2n over 1 plus R at n also equals 0 and you must state by the squeeze theorem all right so it&#39;s very important to remember anytime you apply the squeeze theorem you have to rewrite the original sequence with proper limit notation and state that you used the squeeze theorem all right looks nice moving on to the next idea theorem if the limit as X approaches infinity of f of X so now we&#39;re talking about a real valued function equals L and f of N equals a sub n so that&#39;s our sequence where N is a natural number then the limit as n approaches infinity of f of n is equal to the limit as n approaches infinity of a sub n which is out so what this basically tells us is if we have a function whose infinite limit is L and that function can also be used to represent our sequence but just remember now our domain is going to be the set of natural numbers then they would have the same infinite limit and this is going to be really useful so that we can apply additional limit techniques so let&#39;s consider an example here a sub n is equal to natural log of N squared over N so just at first glance when you&#39;re taking a limit for sequences we always take the limit to infinity and I noticed that the numerator is approaching infinity and the denominator is also approaching infinity as n gets very very large so I have an indeterminate form of the type infinity over infinity and one of our first inclination is to use l&#39;hopital&#39;s rule but remember we&#39;re only allowed to use l&#39;hopital&#39;s rule as long as both the numerator and denominator of our expression are differentiable functions and we know that sequences are definitely not differentiable the domain is a discrete set it&#39;s a set of natural numbers so what I have to do is define a function I&#39;m gonna say let s of X equal and then I&#39;m just gonna copy exactly the same sequence but replace all the ends with X&#39;s so then we&#39;re gonna have the natural log of x squared over X now I&#39;m going to consider the limit as X approaches infinity of this function and I know from my theorem before whatever the limit of this function is it&#39;s going to be the same as the limit of our sequence so this is the limit as X approaches infinity of the natural log of x squared over X again as X gets arbitrarily large the numerators approaching infinity denominators also approaching infinity so I&#39;m going to apply l&#39;hopital&#39;s rule make sure you indicate that you&#39;re doing so so this is gonna be the limit as X approaches infinity derivative of the numerator so I&#39;m gonna use the power rule I have two times the natural log of X and then by the chain rule I have to multiply by the derivative natural log of X which is 1 over x over and then the derivative of the denominator the derivative of X is just 1 all right so I can rewrite this as the limit as X approaches infinity of 2 Ln of X over X all right now let&#39;s investigate what&#39;s going on so as X approaches infinity the numerator is approaching infinity the denominator is also approaching infinity so again I have an indeterminate form of the type infinity over infinity so I can apply l&#39;hopital&#39;s rule again and we have the limit as X approaches infinity derivative of 2 times the natural log of X is just going to be 2 over x over derivative of X is just 1 so now I can rewrite all of this as the limit as X approaches infinity of 2 over X and if I have a constant over the denominators approaching infinity this is equal to 0 so what is this all Tommy well basically now I can say all right based on our theorem the limit as n approaches infinity of a sub n is equal to the limit as n approaches infinity of the natural log of N squared over n which also equals 0 so we would say this sequence converges all right make sure anytime you do this anytime you apply l&#39;hopital&#39;s rule you first define a function f of X otherwise your work is incorrect because l&#39;hopital&#39;s rule is not valid unless you have a differentiable function so you cannot apply it to a sequence directly all right moving on another fact say we have the limit as n approaches infinity of R to the N and R is between negative 1 and 1 then that limit is going to be 0 that should make sense because basically you&#39;re to get from 1 turn to another you&#39;re multiplying by something smaller than one so the sequence is going to be shrinking shrinking shrinking from one term to the next so it&#39;s going to get to zero eventually notice when art is equal to positive one the limit as n approaches infinity of R to the N would just be positive one and that converges also so here we have an example where we&#39;re going to try to apply this we have a sub n our sequence equals 3 to the n plus 2 divided by 5/2 then so let&#39;s see if I can rewrite it so I can identify what are that common ratio is and I&#39;m gonna use my laws of exponents and break up the n plus 2 in the numerator and we write it as 3 to the N times 3 squared over 5 to the N and now I can group 3 to the N over 5 to the N and 3 squared that&#39;s just 9 right this is 9 so rewriting my sequence I have 9 times 3/5 raised to them so if you wanted to list off the first few terms to kind of get a feel for what&#39;s going on you would have 9 times 3/5 to the first and then 9 times 3/5 squared so that would be 9 over 25 and then 9 times 3/5 cubed which is 27 over 125 etc and you could see from one term to the next they&#39;re always going to be shrinking so the sequence is approaching zero about how do we justify that well we&#39;re going to take the limit as n approaches infinity of 3 to the n plus 2 over 5 to the M and we know now we can rewrite that as the limit as n approaches infinity of 9 times 3/5 to that and then using one of our limit laws I can take the 9 out side of the limit so I have 9 times the limit as n approaches infinity of 3/5 to the N since this our value is less than one then I know that that limit goes to zero and you want to indicate that this goes to zero and here&#39;s our justification make sure you include that so if that goes to zero then we have 9 times zero so this limit is zero and our sequence converges all right nice job couple more ideas let&#39;s summarize before we add on so first summary if you see that you have a geometric sequence you have some common ratio raised to a power you want to check if R is between negative 1 and 1 inclusive that&#39;ll let you know if it&#39;s convergent or not squeeze theorem that&#39;s especially useful for sine or cosine in a sequence if you have a sub n divided by B sub n where they&#39;re both polynomials numerator and denominator you can use l&#39;hopital&#39;s rule if you change everything to be a function of X or you can just divide by the highest degree of the denominator now new definitions an increasing sequence is one where the first term a sub 1 is less than a sub 2 is less than a sub 3 for all N greater than or equal to 1 increasing if the terms get larger decreasing sequence if a sub 1 is greater than a sub 2 is greater than a sub 3 for all and greater than or equal to 1 a monotonic sequence is one that&#39;s either increasing or decreasing meaning that all the terms are either going to be increasing for the entire domain or always decreasing one or the other now we say a sequence is bounded above if there&#39;s some number H such that all of the terms are less than H for all N greater than or equal to one and the sequence is bounded below if there&#39;s a number lowercase M such that all a sub ends are greater than or equal to M for all n greater than or equal to one so basically all the terms are bigger than some number a bounded sequence is a sequence that has a upper bound H and the lower bound M so it&#39;s bounded above and below and the most exciting part of all of this is our theorem here that says every bounded monotonic sequence converges and I think the best way for you to really get a feel for that theorem is to look at a little graph so we&#39;re gonna graph some arbitrary sequence but remember it&#39;s bounded right so when we say a sequence is bounded that means it has both an upper bound and a lower bound it doesn&#39;t matter what they are so we&#39;ll say here&#39;s H our upper bound all the terms have to be less than H or equal to it and then our lower bound lower case and I&#39;ll just put it here okay so our sequence a sub n I don&#39;t know what it&#39;s doing but it lives between H and M okay it&#39;s only allowed to live in there and then the other part of the theorem says this sequence is also monotonic meaning it&#39;s either strictly increasing or strictly decreasing so pick a point to start it doesn&#39;t matter where I&#39;m gonna call that one a sub one say it&#39;s strictly decreasing that means I&#39;m only allowed when I move to the next term to decrease well there&#39;s no way for this sequence to not converge or to not have a limit of n approaches infinity if it&#39;s bounded it can&#39;t do something crazy like spiral down towards negative infinity because it&#39;s bounded below by lowercase M what if you started off somewhere house maybe down here and the sequence was increasing well again if it&#39;s strictly increasing it since it has to be monotonic it doesn&#39;t have the option of going back down it can only keep getting larger and larger so the limit would be capital H but notice you need a sequence to be bounded both above and below and be monotonic in order to apply this theorem alright so let&#39;s see how we can apply it consider the following sequence a sub n which equals 1 over n the terms are as follows if you want to list out a few to get a feel for it 1 1/2 1/3 1/4 now a sub n is bounded if I&#39;m gonna make a statement like that I need to find what the upper bound and lower bound are well it&#39;s bounded above by 1 because that&#39;s the largest term that the sequence can ever be and it&#39;s bounded below by 0 right there&#39;s no way for me to get a negative term for a sub n since I&#39;m only substituting in the natural numbers for n also this sequence is monotonic because it&#39;s strictly decreasing this means that it converges based on our theorem in fact we know that the limit as n approaches infinity of a sub n is equal to 0 now what if I didn&#39;t have all three of those conditions say I considered the following sequence a sub n which equals negative 1 to the n plus 1 well let&#39;s list out the first few terms so if I substitute in 1 for n that&#39;s gonna be negative 1 to the 1 plus 1 so negative 1 squared which is 1 then the next term would be negative 1 then 1 then negative 1 etc if a sub n bounded can you find an upper bound in the lower bound absolutely yes it&#39;s bounded it&#39;s bounded above by 1 right it&#39;s never gonna be larger than 1 and it&#39;s bounded below by negative 1 however is it monotonic wanted negative 1 1 negative 1 it switches from increasing to decreasing which means a sub n is not monotonic which means that a sub n does not converge in fact it diverges and if you want to see this this one&#39;s pretty fun one to graph because it&#39;s just gonna be bouncing between 1 and negative 1 right so here&#39;s n here&#39;s a sub n here&#39;s a positive 1 and negative 1 and all its gonna do is just bounce between those two so it&#39;ll be a positive 1 for the first term 1 2 3 4 5 then negative 1 then positive 1 then negative 1 and positive 1 etc so as n approaches infinity you could see the sequence isn&#39;t leveling off its not approaching any finite number alright last example here determine whether the sequence is increasing or decreasing or not monotonic is it banded alright so here we have a sequence a sub n which equals to n minus 3 over 3 n plus 4 well last time we had a really easy sequence just 1 over n we listed up the first few terms and we could clearly see that it was decreasing let&#39;s try listing out the first few terms here and see what happens so a sub 1 if I substitute in 1 for n that&#39;s gonna be 2 minus 3 over 3 plus 4 so that&#39;s negative 1 7 and then a sub 2 that would be 4 minus 3 over 6 plus 4 so that&#39;s one-tenth a sub 3 is 6 minus 3 over 9 plus 4 so that&#39;s gonna be 3 over 13 a sub 4 it&#39;s gonna give us 5 over 16 so it&#39;s not super obvious what&#39;s going on here what&#39;s another way that we could test whether something&#39;s increasing or decreasing well from calc 1 we know that the first derivative gives us information about where a function is increasing or decreasing but again I can&#39;t take a derivative of a sequence it&#39;s not differentiable so similarly to when we wanted to apply l&#39;hopital&#39;s rule i&#39;m gonna consider a function f of X and I&#39;m gonna let it equal to X minus 3 over 3x plus 4 alright so redefine your sequence now of the function this is going to be differentiable so f prime of X is equal to so that&#39;s a need to use the quotient rule I&#39;m gonna have 3x plus 4 times derivative of the numerator that&#39;s gonna be 2 minus 2x minus 3 times derivative of the denominators 3 and then square the denominator alright so cleaning up the numerator we&#39;re gonna have 6x Plus 8 minus if I distribute 3 that&#39;s gonna give me a 6x and then a positive 9 over 3x plus 4 squared and this is it being a really nice derivative because 6x cancels out so I just have 17 over 3x plus 4 squared well what can I say about this derivative the denominators squared so it&#39;s not going to be negative and the numerator is a positive constant so this is greater than 0 for all real numbers well what does this mean about our sequence if the derivative is positive that means my function is increasing strictly increasing since it&#39;s never equal to 0 so my sequence a sub n 2n minus 3 over 3 n plus 4 is increasing you could even say strictly increasing meaning it never is constant all right now let&#39;s go back to the problem and see what else it asks so we determine that the sequence is increasing is it bounded in order to answer that question I have to find the upper and lower bound if I&#39;m unable to do so than I need to show that the sequence either approaches infinity or negative infinity by taking a limit so let&#39;s see the sequence is increasing then that means automatically I have the lower bound so the lower bound is gonna be the first term because the sequence keeps getting larger as you move to the next term so lower bound I got it it&#39;s just gonna be a sub 1 which is negative 1 7 because it&#39;s increasing great what about an upper bound can I find one well let&#39;s consider the limit as n approaches infinity of the sequence of 2 and minus 3 over 3 n plus 4 well here as n approaches infinity I look at the numerator and the denominator this is first degree in the numerator 1st degree in the denominator so I can just divide the leading coefficient and this limit is gonna be 2/3 so that means the sequence is approaching 2/3 as n goes to infinity it will never achieve that value of 2/3 so that&#39;s our upper bound so since I was able to find both an upper bound and a lower bound that means my sequence is indeed bounded alright so be sure anytime you use the derivative to check if a sequence is increasing or decreasing you redefine it as a function and in order to prove or state that it&#39;s bounded you need to find both the upper and lower bounds that concludes section 11 one stay tuned for 11 too

Transcript for:Sequences for Calculus 2 (Section 11.1)

Transcript for:
Sequences for Calculus 2 (Section 11.1)