we will be adding to our list of series tests today with looking at what's called the direct comparison test which I will be abbreviating as DCT and we will also be looking at the limit comparison test lct both of these require that you're looking at terms that are positive greater than zero this specific one the direct comparison test it says that a sub n needs to be smaller or at least equal to B sub n for our lower values and if you know a series converges and you are going to be able to compare it to a sub n then a sub n converges if you pick a series that you know diverges and you compare it then you'll know that B sub n diverges now that all goes back to this statement so think about it so if a sub n is really big then what has to be true for B sub N and B sub n is even bigger that gives you that but if B sub n converges which means it's approaching a smaller number then a sub n is even going to be smaller than that because a sub n is less than B sub n so try to understand the comparisons now you we were going to see in the direct comparison test that we want all a cévennes to be smaller than B sub n but because the convergence of a series is not dependent really on its first several terms you can modify the test to only require that this is true eventually as in at some point that is going to be a true statement we're also going to be looking at the limit comparison test so here you just have to know that all your terms are positive and then most books just give the limit comparison test as this that if you compare the limit of the ratio of the two series and you get L where L is a limit that's finite and positive then the series will agree meaning they either both converge or diverge however a few other books will also state that if you have a limit that equals to zero then they would both converge and the limit is infinity they would both diverge this first case is the one that's used most often and the great thing is it will work even if you accidentally kind of write the fraction upside down whether you put the a on top of the B on top as long as you get that L is a finite and positive limit then they're still going to agree on convergence or divergence however these two cases are special you have to make sure that the fraction is what I would call right-side up in other words the a sub n has to be on top of B sub n for this to work now you're going well what do I know a sub n B sub n these two series is what you're going to choose to compare and how are you going to know what to choose that's the hard part we're just going to choose things and that's why we're going to practice I also want to say that just with the direct comparison test the limit comparison test can be modified to require that the terms be positive at some point as in eventually at the beginning you may have a few negatives but as long as they eventually are all positive the tests will still hold the only way we get better at all of these tests is to practice so let's just dive in so we're going to determine the convergence or divergence clearly state your reasons now remember you can use any test that you think might work but these first few are designed where you're going to have to use either the direct comparison test or the limit comparison test because these are not going to be geometric they're not telescoping they're not a p-series the nth term test fails and I just like to avoid the integral test as all possible all right so let's look at number one and let's think about that comparison test meaning you have to pick something that you know all right pick a series that is known when I say is known like a geometric series and you know what R equals to a p-series where you know what P equals two those are really what you're going to compare it to how do you know well let's look at this look at the powers of n and kind of disregard the numbers what do you think that it would reduce to 1 over N cubed do I know anything about the series of 1 over N cubed yes I do this is a known series this is a p-series where PS 3 which is greater than 1 so I know this converges but I now have to compare it so the direct comparison test what do I have to do well the first thing I have to do is I have to show an inequality so if what I'm picking converges then I'm going to say all right 1 over N cubed converges what I'm comparing it to has to be smaller think about that if this converges to a value and this is smaller than it also has to converge so I have to show that that n equal quality is true how can I do that well one of the easiest ways is to do some cross multiplication but you have to be very careful and keep it in the right inequality order meaning this has to go on the left-hand side so n cube times n minus 1 is that really less than or equal to 1 times n to the 4th plus 2 so let's see I have n to the 4th minus n cubed is that less than or equal to n to the 4th plus 2 is that true yes it is because of that negative so if you take a negative number and cube it you're still going to get something negative which is definitely less than positive 2 so I can then conclude that the series the given series also converges because of the direct comparison test I know that's weird just keep going right so let's look at example number two again I have to pick a something that I know about so I just kind of ignore all the numbers and say well wouldn't this kind of look like 1 over the square root of N and you're going what's that well remember the square root is the 1/2 power that's also a p-series where P is 1/2 which is less than 1 so this diverges that's different that's okay I know that that diverges so I'm going to pick my known series which is 1 over I'm just gonna rewrite that as the square root of N and if that diverges I have to show that that is smaller than my given series because remember I want it to agree so one more time if this diverges and this one is even bigger then it would also diverge so I have to show that inequality so again I'm going to cross multiply being very careful 1 times this querida in is that really greater than or equal to 1 times 2 plus the square root of n is that true well no square root of n is not bigger than 2 plus the square root of n so this is not true did I do anything wrong no I just have to use another test well since the direct comparison test didn't work maybe I should try the limit comparison test ok so the direct comparison test failed so let's try the limit comparison test the neat thing about the limit comparison test I'm going to still use the known series so I'm gonna still use all of that information so I'm going to compare it to I'm just gonna rewrite this again the square root it in where P is 1/2 which is less than 1 which diverges so the limit comparison test so I'm going to take the limit as n approaches infinity what I choose to compare it to is going to go in the denominator if you always do that you really can't go wrong what I'm comparing it to or the given is gonna go on top well that's yup so let's just do some algebra so remember when you divide by a fraction you flip and multiply by the reciprocal so this would be this would come up here and I would have then the square root of n over 2 plus the square root of N and then can I take the limit well sure because these square roots are the same power so I'd the same power over the same power so that limit is just 1 which is good because that is a finite positive number and my limit comparison test says if I have a finite positive number as a limit then these two series agree so I'm going to say therefore the given series diverges because of the limit comparison test okay so I've shown what I've compared it to how I know what I'm comparing it to diverges and then I've taken the limit I've made a conclusion written all my steps down