in this video i'm going to give you a review of the different tests that you need to know in order to determine if a series is going to converge or if it's going to diverge so the first test we're going to talk about is the divergence test now what you need to do is take the limit as n approaches infinity of the sequence a sub n if this limit does not equal zero let's say if it equals five one half infinity if it doesn't equal zero then the series diverges if it equals zero the series may converge or may diverge you need to use another test now the next thing we need to talk about is the geometric series so first you need to identify the form that it comes in and it usually looks something like this there's going to be a constant a and then your common ratio r is typically raised to the n or n minus 1 power but your goal is to identify the value of r so if r that is if the absolute value of r is less than one then the series is going to converge now if the absolute value of r is greater than or equal to one then the series will diverge now the next task you need to be familiar with is the p series tests so the p series is in the form of 1 over n raised to the p so what you need to know is that if p is greater than 1 the series will converge but if p is less than or equal to one then the series is going to diverge next we have the telescoping series now let's say if you have a series where when you write out the terms it's like one minus one over two plus one over two minus one over three plus one over three minus one over four and typically you can cancel the terms in the middle now what you want to do because this will go to the general formula a sub n you want to write a formula that will give you the partial sum of the telescope in series it could be 1 minus a sub n or 1 plus a sub n or something of that nature but once you write the general formula for the partial sum then you want to evaluate the sum of the series as n approaches infinity which is going to be the limit as n goes to infinity of the partial sum formula now if you get some finite value let's call l then that means that the series converges which is usually the case now if you get let's say plus or minus infinity or doesn't exist then the series is divergent so here is a typical example of a telescoping series problem let's say if you have the series from one to infinity of one over n minus one over n plus one that's a telescoping series problem if you list out the terms it's going to be 1 over 1 minus 1 over 2 and then plus 1 over 2 minus 1 over 3 and so forth now sometimes you may not have two separate fractions you may only have a single fraction and you need to use partial fraction decomposition to split it into two separate fractions and then you can list out the terms so watch out for the telescope and series problem you need to keep an eye out for that one now the next test need to be familiar with is the integral tests so what do you recall about the integral test so let's say if we have the sequence a sub n and we're going to make it equal to a function of n f has to be positive continuous and a decreasing function and it has to be decreasing from some number n to infinity so the long term trend is it has to be decreasing sometimes it can go up initially but as long as it's decreasing for the long run it can work so what you need to do is take the integral from one to infinity of f of x dx and let's say if you get some finite value out then that means that the series the original series converges now let's say if you get plus or minus infinity or you get an answer where it doesn't exist then the original series diverges and so that's the integral test the next one we're going to talk about is the ratio test in this case you need to take the limit as n goes to infinity of the absolute value of a sub n plus 1 divided by a sub n now if your result is less than 1 then the series converges if your result is greater than one or if it goes to let's say plus positive infinity negative infinity will apply here because your final answer has to be positive then the series will diverge now if the limit is equal to 1 then the ratio test is inconclusive next up is the root tests so what you need to do is take the limit as n goes to infinity of the nth root of the absolute value of a sub n now if you get an answer that's less than one then the series converges now if you get a value where it's greater than one or if it goes to infinity then the series diverges if your answer is equal to one then like the ratio test the root tests will become inconclusive and so you may need to use another test to get the answer to determine if the series is going to converge or diverge now let's talk about the direct comparison test what do you remember about this test well first we need to have two sequences a sub n and b sub n so let's say b sub n is the big sequence and a sub n is the small sequence and so here's the basic idea behind the direct comparison test so if the big series converges then the small series will converge as well likewise if the small series diverges then the big series diverges as well now let's talk about the limit comparison test this test is very similar to the ratio test the only difference is you're dealing with two different sequences as opposed to one so you need to take the limit as n goes to infinity of a sub n divided by b sub n you need to show that this is equal to some finite positive number which we'll call l and if that's the case then both series will either converge or both series will diverge so let's say if you know that this series converges and if you pass the limit comparison test then you can say that the series with b sub n that's going to converge as well or let's say if this series diverges then you know this series the other one will diverge as well so either they both converge or they both diverge and that's the basic idea behind the limit comparison test if you know the convergence or divergence of one series then you could determine the convergence or divergence of another series next up is the alternating series test and this is when you have a series with alternating signs so typically it's in this form you're going to have negative 1 raised to the n or n plus 1 times some sequence a sub n and in order to determine if the series will converge two conditions must be met the first condition is that it has to pass the divergence test so you need to take the limit as n goes to infinity and you need to show that it's zero if it's not zero then the series diverges if it's zero then you need to move on to step two you need to show that the sequence is decreasing so a sub n has to be equal to a greater than a sub n plus one the next thing you could try is the absolute value test so let's say if the absolute value of the series converges then the original series will also converge and when this happens you can say that the series the original series is absolutely convergent now let's say if you take the absolute value of the series and you find that it diverges and you take you analyze the original series but it doesn't diverge let's say it's convergent if the original series is convergent then the original series is conditionally convergent now let's say if you analyze the absolute value of the series and you find it to be divergent then you check the original series and that's divergent as well then the whole series is simply divergent there's nothing to be said about it so let's say if we have the series from 1 to infinity 2n squared plus 5 over seven n squared minus four so what do you think we need to do for this problem which test should we use if you're not sure it's always good to start with the divergence test so let's take the limit as n goes to infinity of a sub n now a sub n is what we see here so what's the limit equal to if you want to you can use l'hopital's rule so let's take the derivative of the numerator and the denominator so the derivative of 2n squared that's going to be 4n and the derivative of 7n squared is 14n and then what we could do is use l'hopital's rule one more time and so this is going to be 4 over 14 which reduces to 2 over 7. so notice that the limit as n goes to infinity of a sub n it doesn't equal 0. therefore the original series is divergent according to the divergence test and that's it for this problem consider this series let's say if we have the cube root of n divided by n to the fifth power will the series converge or will it diverge so which tests would you use here notice that we can put the series in this form you can simplify it or reduce it to this form so we're dealing with a p series so the cube root of n that's n to the one third and let's divide it by n to the fifth so we need to subtract the exponents what's one over three minus five so getting common denominators let's multiply this by three over three and so this is one over three minus fifteen over 3 so that's going to be negative 14 over 3. so we can rewrite the series like this so we have n to the negative 14 over three which we can rewrite that as one over n to the fourteen over three so now it's in the proper format so we can clearly see that p is equal to fourteen over three and fourteen over three is greater than one so p is greater than one and for the p series if p is greater than one then the series converges now let's try another problem so let's say this is five times one fourth raised to the n minus one what type of series do we have notice that it's in this form so what we have is the geometric series so the common ratio is one over four now recall that if the absolute value of r is less than one the series converges for geometric series and if it's greater than or equal to one the series diverges one-fourth is certainly less than one so we can say that this particular geometric series it converges now what if we wanted to find the sum of the geometric series as it goes to infinity if r is less than one then the sum is equal to the first term divided by one minus the common ratio so to find the first term we need to plug in n equal one so that's going to be five times one over four raised to the one minus one or zero one fourth raised to the zero power is one and so we get five so five is the first term so it's going to be five divided by one minus one fourth so we need to multiply the top and the bottom by four five times four is twenty and then distribute the four four times one is four and four times the fourth is one so the sum is going to be twenty over three so the value of this series is equal to this number now let's try this one so let's say we have negative 1 raised to the n power divided by the square root of n so what type of series are we dealing with here well let's write out the first few terms when n is 1 it's going to be negative 1 divided by the square root of 1 and when n is 2 is going to be negative 1 squared is positive 1 divided by the square root of 2 and then this can be negative 1 over the square root of 3 and so forth so notice that we have an alternating series so what should we do here how can we determine if the alternating series will converge or diverge two conditions must be met first it has to pass the divergence test so if we take the limit as n approaches infinity of a sub n now you need to know what a sub n is in this problem so for the alternating series it's typically in this form it's going to be negative 1 to the n power times a sub n so a sub n is everything except this this could be n plus 1 2 by the way just keep that in mind so a sub n is 1 divided by the square root of n so if we take the limit as n goes to infinity of 1 divided by the square root of n then this is going to go to zero so it passes the divergence test now the second thing that we need to show is that the next term is less than or equal to the previous term so the previous term being 1 divided by the square root of n and the next term is going to be 1 over the square root of n plus 1. so the denominator of this fraction is bigger which means the value of this fraction is smaller so this is a true statement therefore we could say that this series according to the alternating series test is convergent now is it absolutely convergent or is it conditionally convergent what would you say so let's analyze the convergence of the absolute value of the series so if we take the absolute value of negative 1 to the n divided by the square root of n the negative 1 to the n it produces the alternate signs 1 negative 1 negative 1 1 and so forth so if you take the absolute value of that the negative one and the one will just be one so then we're gonna get this series so it's gonna be one divided by the square root of n and the square root of n is n to the one half so what type of series do we have notice that we have the p series it's in the form one over n to the p so we can clearly see that p is one half so if p is one half will the series converge or will it diverge in order for the series to converge p has to be greater than one but because p is less than or equal to one this series is divergent so because the original series converges but the absolute value of the series diverges then we say that the original series is conditionally convergent in order for it to be absolutely convergent the original series must converge and the absolute value of the series must converge as well now let's move on to our next example so let's say this is 1 over n times n plus one will the series converge or diverge well let's check with the divergence tests so what happens when n goes to infinity so we can rewrite this as the limit as n goes to infinity of 1 over n squared plus n and if we multiply the top and the bottom by one over n squared we're going to get the limit as n goes to infinity one over n squared divided by one plus one over n so 1 over n squared that's going to go to 0 and 1 over n is going to go to 0. so it equals 0. so the divergence test is inconclusive so what do you think we need to do here this problem can be solved if we recognize it to be a telescope and series problem so first we need to use partial fraction decomposition so let's set one over n times n plus one to a over n plus b over n plus one and let's multiply both sides by n times n plus one so these will cancel and that's going to give us one and then n will cancel leaving behind a times n plus one and then m plus one will cancel leaving behind b times n now let's say if n is equal to 0 that means b n disappears so 1 is equal to a times 0 plus 1 which means a is equal to 1. now if n is equal to negative 1 this will disappear that will be 0. so 1 is equal to b times negative 1 which means b is negative 1. so we can rewrite this expression like this so instead of one over n times n plus one we can replace a with one so it's going to be one over n and then we can replace b with negative one so it's negative one over n plus one so now you could see that we have a telescope in series so let's write out a few terms so when n is one this is going to be one over one minus one over two so this is the first term and then the second term when n is two is going to be plus one over two minus one over three so that's a sub 2 and then when n is 3 is going to be 1 over 3 minus 1 over 4. so that's the third term now let's write the a sub n minus one term so if we replace this with n minus one it's just going to be one over n minus one and then if we replace this with n minus 1 plus 1 it's just going to be 1 over n now the last term that is the a sub n term is just going to be what you see here so one over n minus one over n plus one so all of this is the sum of the first n terms so this is the partial sum now let's see what we can cancel so we can cancel negative one-half and one-half negative one-third and one-third and then negative one-fourth will cancel with one-fourth which is somewhere in the middle of the sequence and then we can cancel negative one over n plus one over n now one over n minus one will cancel with another one over n minus one if you write out the a sub n minus 2 term so the only thing that's going to remain is 1 and negative 1 over n plus 1. so this allows us to write the general formula for the partial sum so it's a one minus one over n plus one let's get rid of this too so now we need to calculate the sum of the sequence so let's take the limit as n approaches infinity of the partial sum formula s sub n so that's one minus one over n plus one so this is equal to one minus the limit as n goes to infinity of one over n plus one and this is going to go to zero so it's one minus zero which is one so therefore the sum of this series is equal to one and because it's equal to a finite positive number and not infinity the original series we say converges here's another one that you could try let's say if it's one over n squared plus four so what can we do for this problem how can we determine if the series is going to converge or diverge what tests would you recommend the best thing to do is to compare it to use the direct comparison test notice that if we take away the four this becomes a p series problem now which sequence is bigger one over n squared plus four or one over n squared one over n squared is greater than one over n squared plus four if you plug in one this would be one and this is going to be one over five so this is always going to be bigger for all n so therefore we could say this is our a sub n term and this is our b sub n term so this series is bigger than this one now according to the p series we see that p is equal to two we know this series converges because p is greater than one now according to the direct comparison test if the big series converges then the small series will converge as well and that's all we need to do for this problem so this is going to be convergent now let's move on to our next problem so let's say if we have a series which starts from 3 and goes to infinity and it's 1 over the square root of n minus 2. so what sort of tests should we use in order to determine the convergence or the divergence of the series well we can start with the divergence test but when n becomes very large because it's bottom heavy the limit will go to zero and so based on the divergence test it may diverge or may converge next we could try to derive comparison tests because if we get rid of the two then this becomes a p series and the square root of n is n to the one half which means that p is one half so because p is less than or equal to one this series diverges now which series is bigger is it this one or this one because n minus 2 is less than n we have a smaller denominator if you decrease the value of the denominator of a fraction the value of the whole fraction goes down i mean excuse me the value of the whole fraction goes up so therefore this series is larger so we should write greater than or equal to so this is going to be the big series b sub n and this is going to be the small series a sub n so if the small series diverges will the big series diverge as well and the answer is yes the big series must diverge as well so this could work now there's another test that you could use if you want to and that is the integral test so let's say f of x is one divided by the square root of x minus two now the function is not continuous at x equals two because that's a vertical asymptote but we're starting from three so it is continuous from three to infinity we need to show that it's continuous decreasing and positive this is always going to be positive especially when x is greater than 3. and because we have a square root of x minus 2 on the bottom it's going to be a decreasing function as x increases the value of the denominator will go up and so the function will decrease now you can also find the first derivative and show that it's negative on this interval if you do that that tells you that the function is decreasing so now once those conditions are met once you show that it's positive continuous and decreasing on this interval take the integral in this case starting from 3 to infinity sometimes it's usually 1 to infinity integrate f of x dx from a to b so what we have here is an improper integral so first i'm going to find the indefinite integral i'm going to use u substitution so if i make u equal to x minus 2 d u is going to be equal to dx and so i can replace x minus 2 with u and dx with du so i can rewrite this as u to negative one half and so that becomes u to the positive one half divided by one half or multiplied by two and so this becomes two times the square root of x minus two so this is equal to the limit as a goes to infinity of 2 times the square root of x minus 2 evaluated from 3 to a so now let's plug in a and three so we're going to have the limit as a goes into infinity and this is going to be 2 times the square root of a minus 2 plus two times i mean plus but minus minus two times the square root of three minus two now as a goes to infinity this whole expression becomes infinity and so infinity minus 2 times 1 is just infinity so if the integral test gives you infinity that means that the original series diverges which was in harmony with the other conclusion that we got based on the direct comparison tests let's say if we have the series of the square root of n divided by n cubed plus two which test should we use for this one the test that i recommend using is the limit comparison tests but we need two separate functions one of which is more simplified than the other so for the other series i'm just going to get rid of the two so it becomes the square root of n divided by n cubed and according to the limit comparison test if we take the limit as n goes to infinity of the ratio of the two sequences if it's equal to a finite positive number then both series will either converge or they will both diverge now we can reduce this to a p series so this is n to the one half divided by n cubed and so a half minus 3 is 2.5 negative 2.5 and so this becomes 1 over n to the 2.5 so p is 2.5 and that's greater than one so according to the p series test this series converges so if this passes the limit comparison test it will also converge let's find out now let's take the limit as n goes to infinity of a sub n divided by b sub n so let's call this our a sub n and this expression is going to be our b sub n sequence so this is going to be the limit as n goes to infinity and then a sub n that's the square root of n divided by n to the third plus 2 and then divided by b sub n which is the square root of n over n cubed perhaps you heard of the expression keep change flip keep the first fraction the same change division multiplication flip the second fraction so we can rewrite this as times n to the third divided by the square root of n so these two will cancel and so we have the limit as n goes to infinity of n cubed divided by n cubed plus two so let's multiply the top and the bottom by 1 over n cubed so if we do that we're going to get the limit as n approaches infinity and this is going to be 1 divided by 1 plus 2 times 1 over n cubed so when n becomes very large 1 over n cubed goes to 0. so it's 1 over 1 plus 2 times 0 which is 1 over 1 and that's 1. so this is equal to a finite positive number therefore according to the limit comparison test because this series converges this one must converge as well now consider the next example let's say it's 3n squared minus nine divided by seven n squared plus four raised to the n which tests should we use if you see something raised to the n power it's a good indication that we need to use the root test so recall that the root test is this take the limit as n goes to infinity of the nth root of the absolute value of the sequence a sub n now if it's less than one the series converges if it's greater than one or if it's infinity the series diverges if it's equal to one then it's inconclusive so if we take the limit as n goes into infinity of the nth root which is the same as raising everything to the 1 over n power and times 1 over n those will cancel and so we have the limit as n goes to infinity of three n squared minus nine divided by seven n squared plus four notice that the degree of the numerator is the same as that of the denominator so you can see that the limit is going to be 3 over 7. but if you want to show your work you can multiply the top and the bottom by 1 over n squared or use l'hopital's rule let's use l'hopital's rule for this example so the derivative of 3n squared is 6m and the derivative of 7n squared is 14n so using l'hopital's rule again this is going to become six over fourteen which reduces to three over seven so this is the answer now three over seven is less than one so based on the root tests the series converges now let's work on our final example and so we're going to have the series of 2 raised to the n divided by n factorial so what tests should we use for this example so since he knows the last problem it's going to be the test that we haven't used yet which is the ratio test and so the basic idea behind the ratio test is we need to take the a sub n plus 1 and divided by a sub n all within an absolute value symbol and if this is less than one the series converges if it's greater than one or equals infinity the series diverges now if it's equal to one then it's inconclusive now we know that a sub n is 2 to the n over n factorial but what's a sub n plus 1 all you need to do is replace n with n plus one so the limit as n approaches infinity is going to be a sub n and then divided by i mean i'll take that back it's a sub n plus one we need to write that one first and then divided by a sub n using keep change flip we could say it's multiplied by one over a sub n or the reciprocal of a sub n so this is a sub n plus one times one over a sub n so you can represent it that way if you want to so how can we simplify the limit now what do we need to do 2 to the n plus 1 that's two to the n times two to the first power n plus one factorial is n plus one times n factorial and so we can cancel 2 to the n and we can cancel n factorial so we're left with the limit as n goes to infinity of 2 divided by n plus one two over n plus one is going to go to zero as n goes to infinity so zero is less than one which means that the series converges by the ratio test and that's the answer you