 We will continue our discussion about the infinite series, which we started in the last class. Let us recall that infinite series is an expression of the following form sigma x n, n going from either 1 or 0 or some such number. And we have seen that along with every such infinite series, we associate a sequence s n, which we call a sequence of partial sum. It is sum of the first n elements in that sequence. And depending on whatever happens to s n, we say that if s n converges, we say that the series converges or summable. And whatever is the limit of s n, we say that is the sum of the series. And similarly, if s n diverges, we say that the series diverges. And so we can easily see the, let us see one or two standard examples of this before proceeding. So, let us, now this is, as you know it is called geometric series sigma x to the power n. And it is customary to start this for n going from 0 to infinity. So, in this case, if you look at s n, s n will be 1 plus x plus etcetera, etcetera up to x to the power n, x to the power n. And so this is nothing but since this is a geometric progression, so you can sum this as 1 minus x to the power n plus 1 divided by 1 minus x. And so, whether the series converges or not depends on whether the limit of this exists as n goes to infinity. And out of you can see, only term which depends on n here is this, x to the power n plus 1, all the terms, other terms are independent of n. And here we have seen that this converges to 0, if mod x is less than 1. We have seen that this converges to 0, if mod x is less than 1. So, it means that this series converges to 1 by 1 minus x, if mod x is less than 1. So, we can say that the sum of this series is 1 by 1 minus x, if mod x is less than 1. If mod x is bigger than or equal to 1, this limit will not exist, limit of this will not exist and this series diverges, that is also fairly easy to see. Let us see one more example. Suppose you take sigma 1 by n into n plus 1, n going from 1 to infinity. Now, here only worth file thing to notice is the following, that 1 by n into n plus 1 is nothing but 1 by n minus 1 by n plus 1. And hence, if I look at this partial sum S n, S n will be, let us write S n as x 1 plus extra x n. So, that is x 1 is 1 minus 1 minus 1 minus 1 minus half, that is 1 by n minus 1 by n plus 1. So, x 2 is 1 by 2 minus 1 by 3 etcetera. So, finally x n is 1 by n minus 1 by n plus 1. And it is clear that middle terms will cancel and the final sum will be nothing but 1 minus 1 by n plus 1. And this tends to 1 as n minus n tends to infinity. So, this series converges and its sum is 1. So, in general this is how you decide whether a series converges or not. Look at the partial sum and take the limit of the partial sum as n goes to infinity. Next, what we want to see is that apply the various theorems that we have known about the sequences to get the corresponding results about the series. Let us look at that last thing that we saw about the sequences first. We have seen that sequence of real numbers converges if and only if it is Cauchy. It is a Cauchy sequence. So, how does it translate in terms of infinite series? We can say that this, let us write this as theorem. Sigma x n n going from 1 to infinity converges if and only if this s n is a Cauchy sequence. Let us write this in the full form. What has been by Cauchy sequence? That for every epsilon bigger than 0 there exists n 0 etcetera. So, that means if and only for every epsilon bigger than n 0 there exists n 0 in n. Such that for all n and m bigger than or equal to n 0 we must have mod s n minus s n less than epsilon mod s n minus s n less than epsilon. Only thing is I want to make a slight change in the way of writing this mod s n minus s n. n and m both are bigger than or equal to n 0. So, let us say one of them is bigger than the other. So, suppose let us say n is bigger than m. Let us say n is bigger than m and m is bigger than or equal to n 0. Then s n is x 1 plus x 2 etcetera plus x n and what will be s m? s m will be x 1 plus x 2 plus x m then plus if since m is smaller plus x m plus 1 etcetera up to x n. So, what is s n minus s n or s n minus what is s n minus s n? It is it is this part x m plus 1 etcetera up to x n. So, I can say that instead of writing like this I shall say this is same as sigma x j j going from m plus 1 to n. So, if you take any two numbers m and n bigger than n 0 and suppose m is strictly less than n and if you take all the sums of all this sum of all the terms going from x m plus 1 to x n then that should be less than epsilon. In other words given any epsilon for large values of n sum of that part whatever whatever large numbers m and n you take you take the sum from x m plus 1 to x n its absolute value should be less than epsilon. There is one very important and immediate but simple consequence of this. You can see that there is nothing to prove here. This simply follows by observing that we have just rewritten the same thing that s n converges if and only if s n is a Cauchy sequence in a different language. One obvious corollary of this is the following. Suppose I take m is equal to n suppose I take m is equal to n then there will be only one term here and that will be mod x n less than epsilon. So, what is the meaning of that? For every epsilon there exists n 0 such that whenever n is bigger than or equal to n 0 mod x n is less than epsilon. What is the meaning of that? It means x n tends to 0. It simply means x n tends to 0. So, if the series converges if the series if sigma x n converges then x n goes to 0. So, if sigma x n converges and that follows simply by taking n is equal to m here because that is how you define saying that x n converges to 0. So, if sigma x n converges let us write it like this limit of x n as n tends to infinity that is equal to 0. Now, there is a very obvious question here and that is what about the converse? Again it is well known that the converse is false, but we shall and the only way to say show that something is false you need to give a counter example. So, that is something we shall see little later. So, that is about the this is sometimes called Cauchy criteria for the convergence of the series. There is one more thing that follows from this and that is the following. Let us define what is meant by absolute convergence. We say that the series sigma x n converges absolutely n going from 1 to infinity converges absolutely if you take this series of absolute value sigma mod x n n going from 1 to infinity sigma mod x n n going from 1 to infinity that converges. So, again the obvious question is what is the relationship between the convergence and absolute convergence and then again the well known thing here is that every absolutely convergent series is convergent. So, let us see how does it that follow. Now, since we are dealing with two series here sigma x n and sigma mod x n. Let us use some notations for the partial sums of both for this anyway we already derived by s n. So, let us say s n as x 1 plus that should up to x n and suppose partial sum of this series I call t n t n is let us say mod x 1 plus mod x 2 etcetera plus mod x n. We want to show that this series converges we want to show that this series converges and using this theorem it is enough to show that this happens. So, let us just show this let us just show this. So, let us say that let epsilon be bigger than 0 let epsilon be bigger than 0 then we know that since this series converges we know that we have since this series converges this series converges we know that this series satisfies this criteria. So, for that series there exist n 0 such that all these things happen. So, there exist n 0 since since sigma mod x n converges sigma mod x n converges there exist n 0 in n such that for all let us write it n bigger than m bigger than or equal to n 0 mod t n minus t m is less than x 0. But, what is t n minus t m t n is mod x 1 plus mod x 2 plus mod x n. So, similarly we can say that t n minus t m should be same as sigma mod x j sigma mod x j j going from m plus 1 to n. So, that is that is this is nothing but that is we can say that sigma mod x j j going from m plus 1 to n less than epsilon. But, now what is the relationship between this term and this term is it clear that this is less than or equal to this. So, this implies that sigma x j j going from m plus 1 to n absolute value of this that is less than or equal to this and that is less than epsilon. So, this is a very important theorem again let us see what exactly this depends on every absolutely convergent series converges how did we prove this we proved this using this Cauchy's criteria. How was Cauchy's criteria proved by using the fact that every Cauchy's sequence is convergent and how was that proved by using l u b action order completeness of r. So, the fact that every absolutely convergent series converges depends in fact on the completeness or order completeness of the real number system or actually the so called l u b action. Now, since again here also the same question what can we say about the converse that is is it true that whenever a series converges it also converges absolutely that is false and again to show that it is false we need to give a counter example, but that also we shall discuss little later that is when we when we come across some more methods of checking convergence and divergence. Now, but there is one very good thing here is that this gives us a class of convergent series. Let us whatever the series is given what you can do first is to check whether it is absolutely convergent and to check absolutely convergent you just need to look at sigma mod x and the terms of this series are always non-negative. So, if we if we divide some methods of checking the convergence or divergence of the series of non-negative terms that will be good for checking absolute convergence and since we know that every absolutely convergent series converges such series will converge. So, let us first confine our attention for some time to what are called series of non-negative terms. Obviously, what does this mean that we will say that this series sigma x n n going from point to infinity is the series of non-negative term. This simply means that each of x n is bigger than or equal to 0 x n is bigger than or equal to 0 for all. Of course, sometimes these are also called series of positive terms which would mean that each x n is bigger than 0, but you can see that that is a minor difference. If some of the x n's are 0 you can simply remove them from the series it will not make any difference to the series or to the sum of the series. Now, what is the major advantage of this series or over the other series? It is again if you look at the sequence of partial sums. Let us again look at the sequence of partial sums s n. Remember s n is sigma x j, j going from 1 to n and s n plus 1 is this plus x separate s n plus 1 is s n plus x n plus 1 and we know that each x n is bigger than or equal to 0. So, what does it say about the sequence s n? It is a monotonically increasing sequence. Sequence s n if the series of non-negative term then the sequence of partial sums is a monotonically increasing sequence. So, we know this that s n is a monotonically increasing sequence and in case of monotonically increasing sequences we know that converges if and only if it is bounded above. A monotonically increasing sequence converges if and only if it is bounded above. So, that immediately gives us one method of checking convergence of this series. What is that? The series of positive non-negative terms converges if and only if its sequence of partial sums is bounded above. So, we can just write that as a theorem. The series of non-negative terms converges if and only if its sequence of partial sums is bounded above. So, this gives a method of checking whether a series of non-negative terms is convergent or not. All that you need to do is look at the sequence of partial sums and simply show that it is bounded above. If you show that, that is enough. It is enough to show that a series is convergent. Similarly, if you show that it is unbounded then you can immediately conclude that it is a divergent. Now, there is one more thing. The sequence is monotonically increasing. So, even if you are able to show this for some subsequence it is enough. Is that clear? Suppose, instead of showing that S n is bounded above, I simply show that the sequence S 1, S 3, S 5, S 7 that is bounded above. Will that be sufficient? Because S 4 is less than S 5. S 6 is less than S 5. So, even if S 1, S 3, S 5 that is bounded that is enough. Similarly, if that is unbounded again that is enough. So, remember when the sequence is monotonic to show that it is bounded above, it is just sufficient to show that any subsequence is bounded above. Of course, to show that it is unbounded that is always there. Even if you show that some subsequence because if the sequence is bounded all its subsequences also must be bounded. That is true for any sequences whether monotonic or not. So, to show that a sequence is unbounded it is enough to show that some subsequence is unbounded. But in case of monotonically increasing sequences to show that sequence is bounded above for that also it is sufficient to show that some subsequence is bounded above. Now, this particular thing that is this observation leads to again very important observation. The so called comparison test I mean the idea of the comparison test is basically just this that see suppose you are taking two basically here we want to check that the sequence S n once the two once there are two series and if the terms of both are non negative then suppose let us say there are two series x n and y n and suppose for each n x n is less than or equal to y n x n is less than or equal to y n. Then suppose you denote the corresponding partial sums by S n and T n then obviously S n is less than or equal to T n. So, let us just write this suppose there are two series sigma x n n going from 1 to infinity and sigma y n n going from 1 to infinity and suppose this happens suppose 0 less than or equal to x n less than or equal to y n. Now, so what the comparison test says is that if sigma y n converges that sigma x n also converges what is the proof? Proof is what I said just now. So, suppose you take S n as the sequence of partial sums of this x n. So, let us say S n is sigma x j j going from 1 to n and T n as sigma y j j going from 1 to n. We know that x n is less than or equal to y n for each that is in other words x j is less than or equal to y j for each j. So, what does it say about S n and T n? So, it means S n is less than or equal to T n S n is less than or equal to T n remember both are both are increasing sequences. So, all that we need to once we have a series of non-degree terms all that we need to check is that the sequence of partial sums is bounded above. Now, if sigma y n converges it means T n is bounded above. If sigma y n converges it means T n is bounded above. Does it say immediately that S n also must be bounded above? That is it well sometimes this whole thing is also said you know slight I mean let us write that because it is fairly common. If sigma x n diverges sigma y n also diverges can you see that this is basically the same thing written a different language. Because if this were false what must happen that sigma y n converges and sigma sorry sigma x n converges and sigma y n diverges. So, that is not possible if the sometimes this series when this happens we say that the series y n dominates the series x n. And if the series which dominates converges then the series which is dominated also converges. And if the series which is dominated diverges this series also diverges. Right after defining the convergence and divergence of a series I have said that if you remember that convergence or divergence of a series does not depend on what happens to a finite number of terms. So, if a finite number of terms are added or removed it does not matter for the convergence or divergence. Now, using that what we can say here is that see generally we say that this happens for all n. But even if this happens for all n bigger than or equal to some n 0 still it is. Because what happens to the first finite number of terms does not matter. So, in comparison test that is in general we use this condition 0 less than or equal to x n less than or equal to y n for all n. But since it does not matter what happens to a finite number of terms even if this condition is satisfied after some n. Let us say n bigger than or equal to 100 still the same conclusion will be valid. Now, this is a very useful test to check given the series of positive terms converges or diverges. But to make effective use of this test we should have some standard list of series which converges or diverges. And that list is given by this sigma 1 by n to the power p n going from 1 to infinity. And p is bigger than 0 and again this it is a well known result that this will converge if p is bigger than 1. And this diverges if p is bigger than 1 and diverges if 0 less than p less than or equal to 1. So, once we have this list of series given a series of positive terms if it is possible to compare that series with one of these series then we can conclude about the convergence or divergence of any of those series. Let us first look at the proof of this. By the way given a series of or any series not only series of positive terms any series in general there can be several ways of showing that the series converges or diverges not a unique method. For example, we have seen that this series sigma 1 by n into n plus 1 we have seen that this series converges. Now, I can say that we have seen that by different method but this is also series of non negative terms. So, we can also compare this with the series sigma 1 by n square can you see that one this each of this term is less than or equal to each of this term here. And by suppose we prove this then by this it will mean that this is a convergent series. So, that will give this is also convergent series. Of course, this comparison test will not say anything about the sum again there is also not very correct to say it will say that the sum of this is less than or equal to sum of this. Suppose, in this case suppose sigma by n converges and then sigma x n also converges the sum of this series will be less than or equal to sum of this series, but beyond that we cannot say anything. Now, let us see how this can be proved what I will do is that I shall just take one or one sample case of each type one for the convergence and one for divergence and the general case I shall leave it to you as an exercise. So, let me take this case first suppose I take the case p equal to 1 that means we are looking at the series sigma 1 by n going from 1 to infinity we to show that this is divergent we need to show that its sequence of partial sums is unbounded and I told you just sometime ago that it is sufficient to show that some subsequence is unbounded and the idea is as follows you look at this partial sums 1 by 1 by 2 plus 1 by 3 plus 1 by 4 etcetera. Let us say there is some that last point I will write little later you can say that suppose I take the partial sum s 1 that is 1 partial sum s 2 is 1 plus half partial sum s 2 suppose I look at s 4 suppose I look at s 4 then I will say that s 4 is bigger than or equal to 1 plus half plus these two terms 1 by 3 plus 1 by 4. Now, each of these terms are this 1 by 4 is smaller. So, I can say that this is bigger than 2 into 1 by 4 this is bigger than 2 into 1 by 4. So, which is again 1 by 4. So, I can so s 4 is bigger than or equal to 1 plus let us say 1 plus for the time being I should write 2 into 1 by 2 s 4 is bigger than or equal to. Now, next I consider let us say s 8 that is I take 4 next 4 terms that is 1 by 5 plus 1 by 6 plus 1 by 7 plus 1 by 8 take this group. In this group this smallest term is 1 by 8 and suppose I have taken 4 terms the sum of them must be less than sorry bigger than or equal to 4 into 1 by 8. So, that is again 1 by 2. So, that gets added to this because s 4 is already bigger than or equal to 1 plus 2 into 1 by 2 s 8 should be bigger than or equal to 1 plus 3 into 1 by 2. So, now what is the can you predict something from here. Similarly, for example now next suppose I take now s 16 that will be again 1 by 9 plus 1 by 10 up to 1 by 16 the smallest term there will be 1 by 16 there will be 8 such terms there is some is again 1 by 2. So, s 16 will be bigger than or equal to 1 plus 4 into 1 by 2. So, what is the general pattern s s n square not n square 8 is not n square s s 2 power n s that is right s. So, s suffix 2 power n is bigger than or equal to 1 plus n into 1 by 2 and does it follow from here that this s 2 power n is unbounded because it is bigger than or equal to 1 plus n by 2 and this number I can make arbitrarily big this number can be made arbitrarily big. So, this implies that this s suffix 2 power n is unbounded and hence s n also is unbounded and. So, this series diverges and the same proof will work if you take instead of n here if you take n power p for p less than or equal to 1 small modification of this will work to show that the series diverges for this case 0 less than p less than or equal to 1 now let us take 1 from this group p bigger than but let me take sigma 1 by n square and again I say whatever we do for p equal to 2 with obvious small modifications will work for any p bigger than 1 that part I shall leave it to you as an exercise. So, let us look at now again again the idea is basically similar just make the groups of the partial sums here just make the groups of partial group the terms in the partial sum and here we have chosen the smallest among them and then use the inequality less than or equal to now we want to show that s n is bounded. So, we need to use the inequality less than or equal to. So, obviously we should choose the biggest among the terms that is the idea. So, now in this case now s n is 1 plus 1 by 2 square plus 1 by 3 square plus 1 by 4 square etcetera let me begin with s 3 s 3 is 1 plus now I will take the group of these two I will take the group of these two now in these two 1 by 2 square is the bigger term 1 by 2 square is a bigger term. So, the sum so 1 by 3 so the sum of these two must be less than or 2 times 1 by 2 square it must be less than or 2 times. So, this will be less than or say 2 times 2 square next what I will do is I will take again the next four terms that is 1 by 4 square plus 1 by 5 square plus 1 by 6 square plus 1 by 7 square. So, that is that is s 7 not equal to this is less than or equal to. So, s 7 will be less than or equal to 1 plus this 2 by 2 square let me write now as 1 by 2 here the smallest term is 1 by 4 square and there are four such terms. So, the sum should be less than or equal to 4 by 4 square. So, less than 1 by 4 4 by 4 square and now you can predict what is going to happen next I will take next 8 terms that is starting from 1 by 8 square and going up to 1 by 15 square there are 8 terms each is less than or equal to 1 by 8 square. So, the sum should be less than or equal to 1 by 8 and that is s 15. So, s 15 is less than or equal to 1 plus 1 by 2 plus 1 by 4 plus 1 by 8. So, what follows from here again you can predict what will be the general pattern this is number something like here we had got s 2 power n this is not 2 power n it is 2 power n minus 1 it is 2 power n minus 1 and that is less than or equal to you can say on the right hand side you have a geometric series with this common ratio 1 by 2 and we already showed that that is a convergent series. So, these partial sums are bounded. So, what it shows is that this s 2 power n minus 1 this is bounded above see in general for an arbitrary sequence if you show that some subsequence is bounded above will not imply that the whole sequence is bounded above. But, in case of monotonically increasing sequences that is true because what we have shown is that this sequence this subsequence that is s 2 power n minus 1 that is the sequence s 1 s 3 s 7 s 15 that sequence is bounded above. But what we know is that if you take for example s 10 s 10 is less than or equal to s 15 or similarly you should take any n that n is less than or equal to 2 to the power n minus 1. So, s n is less than or equal to s 2 to the power n minus 1. So, the whole sequence also is bounded above and so that shows that this series is convergent. Now, in a similar way you can take any p bigger than 1 and then this same idea will work I shall do that to you as an exercise. So, that is about the comparison test and how one uses comparison test here we have seen an example. Let us now look at one more test for the checking the convergence of this series of non-negative terms that is called root test very powerful test this one. Of course comparison test is also quite powerful suppose this is a series of non-negative terms once again n going from 1 to infinity each x n is bigger than or equal to 0. So, you look at nth root of x n or x n to the power 1 by n whatever you call nth root of x n and consider the limit superior of that. See x n in general nth root of x n may or may not be a convergent sequence if it is bounded limit superior will always exist limit superior will always exist if limit superior is does not exist as a finite real number you take that as infinity. So, let this number be l then what does the test say the test says that if this is the first thing if l is less than 1 then sigma x n converges second thing is that if l is bigger than 1 then sigma x n diverges if l is equal to 1 the test gives no information if l is equal to 1 the test you cannot conclude anything no conclusion no information no information about the convergence or divergence. As far as this last thing is concerned what does it mean that l is equal to 1 it is also possible that the series converges it is also possible that the series diverges and you can see that we already have examples of both this case if you look at this series sigma 1 by n in this case what will be nth root of x n it will be 1 by nth root of n it will be 1 by and we already shown that this nth root of n goes to 0 goes to 1 as n goes to infinity. So, this tends to 1 so in this case l is equal to 1 and the series diverges on the other hand if you look at this series 1 by n square there it will be 1 by n to the power 2 to the power n which will also be its limit also will be 1 and that series converges. So, if l is equal to 1 you cannot conclude anything from the root test now let us look at this one remember that when we discuss this limit superior we have shown that if you take any epsilon if you take any epsilon then there exist some n 0 such that all elements of the sequence x n lie between not lie between or less than that l plus suppose I take any epsilon and then if I look at a number l plus epsilon then I can find a number n 0 such that all the elements of the sequence are less than that l plus epsilon. Now, suppose l is less than 1 I can always choose an epsilon in such a way that l plus epsilon is also less than 1 l is a real number l is less than 1 can I always choose a positive epsilon such that l plus epsilon is also less than 1 obvious choice is 1 minus l by 2 that is a positive number anyway it does not matter you choose some number which is strictly bigger than l and less than 1. Suppose, I call that number eta let the proof of this let l less than eta less than 1 then take the difference between eta and l as epsilon then there should exist some n 0 such that for n bigger than or equal to n 0 x n is less than l plus epsilon that is eta. So, I will simply say that there exist then there exist n 0 in n such that n 0 in n such that x n not x n it is not limit speed of x n remember it is limit speed of n through to of x n it is limit speed of n through to of x n. So, let me write x n to the power 1 by n x n to the power 1 by n that is less than l that is less than not l that is less than it is less than eta l plus epsilon or which is same as saying that which is same as saying that x n is less than eta to the power n x n is less than eta to the power n. Now, you just use the comparison test see you have the series sigma x n and you take this series sigma eta power n you take the series sigma eta power n and this is where I said that this x n less than this is this is not happening for all n it is happening for all n bigger than or equal to n 0, but that is this is something I commented right after discussing the comparison test that is that if x n less than or equal to y n even if it happens for some n bigger than or equal to n 0 still the conclusions of the comparison test are valid. Now, this is a geometric series and it this eta is less than 1. So, this is a convergent series. So, by comparison test this original series should also converge by comparison test the original series should also converge. See you may have learnt this root test even in your under graduate course, but it is highly unlikely that you would have used this limit superior. In the under graduate books the way it means that limit of this n root of x n is equal to l, but this is more general because that limit may or may not exist, but limit superior will always exist limit superior will always exist. So, this you can apply for more general series. Now, let us look at this second part that is also very similar basically follows by using the properties of limit superior. Remember what we have shown is that see this is a limit superior of this sequence. If you remember we all we have shown that there exists a subsequence of this which converges to the limit superior. So, let us suppose I call that subsequence let us say x n j. So, there exists a subsequence x n j converging to l and l is bigger than 1. Now, let me just tell you rough argument here if some sequence converges to something l bigger than 1 is it clear to you that the infinitely many terms of that sequence also must be bigger than 1. Now, if that is the case can that sequence tend to 0 not x n j x n j to the power 1 by n j. So, infinitely many terms of sequence x n j by 1 to the power 1 by n j those are bigger than 1. If this is the case can the sequence x n j tend to 0 it cannot tend to 0 then can x n tend to 0 that cannot happen. If a sequence tends to 0 it cannot happen that infinitely many terms the sequence are bigger than 1, but if the sequence x n does not tend to 0 what do you know about the convergence of the series that was one of our first theorem that if the sigma x n converges then the sequence x n must go to 0 that is not happening here that is not happening here. So, the series sigma x n must be a divergent series and the third case any way we have already seen. So, that is an information about the root test there is one more test of the similar type which is called ratio test that we shall discuss in the next class.