 So, we were discussing the series of non-negative terms in the last class and towards that we saw this root test. Let me just recall what we suppose this sigma a n is the series then we looked at nth root of a n and limb soup of this. And what we said is that if this is bigger than 1 sorry if this is less than 1 the series converges. If this is bigger than 1 the series diverges and if this is equal to 1 this test gives no information. And we also seen examples of both the both the types. Now in this connection there is one more thing that is required when we discuss this series of non-negative terms till now we have been talking about real numbers and real number systems. But for the discussion of this convergence or divergence of a series it is also convenient to discuss what is called extended real line extended real line extended real line means nothing but your usual real line along with these two symbols plus infinity and minus infinity some books denote it as r star this usual real line r star. And these two additional symbols with the usual properties associated with them for example properties like x plus infinity is infinity for every x in r etc. Similarly x minus infinity is minus infinity for every x in r and x multiplied by infinity that is infinity if x is bigger than 0 and minus infinity if x is less than 0. And if x equal to 0 that is usually left undefined and sometimes it is taken as 0 that you will see when you learn major theory in most of the major theory course 0 into infinity is taken as 0. But in general it is left undefined and x divided by infinity that is taken to be 0 and x divided by 0 that is taken as infinity. And we avoid things like just as 0 into infinity is not defined similarly infinity things like infinity minus infinity or infinity by infinity these things are left undefined. That is about the algebraic operations and about the order this is the thing that is for every real number x minus infinity is strictly less than x and strictly less than plus infinity for all x in r. Now this has some convenience for example we have seen that if a monotonically increasing sequence is bounded above then it converges. But if we are using extended real line we can say that if a sequence is protonically increasing then it either converges or diverges to infinity if it is not bounded above then we say it diverges to infinity. Similarly for monotonically decreasing sequence if it is bounded below then it converges to some real number otherwise we say it diverges to minus infinity. In fact some books also use the rotation converges to plus infinity or minus infinity but we shall not use that. Now what are the relevance of this to this root test it is that this l this l can be infinity so that comes under the case l bigger than 1 that comes under the case l bigger than 1 and then so even if this l so for example if you look at the definition of limit superior we had taken the sequence let us say alpha k where alpha k was a supremum of a n for n bigger than or equal to k. So that limit of that sequence can be infinity so limit superior of the sequences can be infinity limit inferior can be minus infinity so even those cases can be included with this convention. Then let us similarly look at the another popular test namely ratio test so here also we are looking at the series let us say sigma a n and each a n is bigger than or equal to 0. So we look at limit superior of a n plus 1 by n that is why it is called ratio test limit superior of a n plus 1 by n so this is the let us say that if this is less than 1 then sigma a n converges if so instead of saying that this limit superior is bigger than 1 I will say is the following if a n plus 1 by a n if this is bigger than or equal to 1 let us say if this happens for all n bigger than or equal to some n 0 then sigma a n diverges and if it we know that for any sequence limit inferior is less than or equal to limit superior. So you look at this sequence so a n plus 1 by a n so suppose the following thing happens that limit inferior of a n plus 1 by a n suppose this is less than or equal to 1 this is less than or equal to 1 and limit superior is bigger than or equal to 1 then the test gives no information then no information that means the series may converge or series may diverge we can give the examples of both the kinds we can say that as far as the proofs are concerned let us dispose of certain things which are trivial as far as this last thing is concerned we can say that we already seen the examples of this for example if you look at say sigma 1 by n and let us say sigma 1 by n square then a n plus 1 by a n will be either something like n plus 1 by n or n plus 1 square by n square and a limit of that will be 1 and once the limit is 1 limit inferior limit superior everything is 1 so it will be this case and we know that this series converges and this diverges so there is nothing new here so the so this in this case this test gives no information similarly if you look at this part 2 if a n plus 1 is bigger than or equal to a n for all n bigger than or equal to n 0 then it will mean that in particular for example a n a n 0 plus 1 is bigger than or equal to a n 0 and then a n 0 plus 2 is bigger than or that means all a n's will be bigger than or equal to a n 0 for n bigger than or equal to n 0 so in which case a n cannot tend to 0 since all of them are bigger than or equal to a n 0 and a n 0 is strictly bigger than 0 so the sequence a n cannot tend to 0 and hence the series cannot converge now let us look at this first case if limit superior of a n plus 1 by a n is less than 1 we can suppose that limit superior is l so let us say l is let us say l is limit superior a n plus 1 by a n and if l is less than 1 then as we have done in the previous case we can always find some number beta which lies between l and 1 so consider some number beta which lies between l and 1 and then we can say that there will exist some n 0 such that for all n bigger than or equal to n 0 a n plus 1 by a n is less than beta because l is limit superior in fact exactly the same argument we use even for the even for the proof of the similar case in the root test also so we can say that there exists n 0 in n such that a n plus 1 by a n is less than or equal to beta for all n bigger than or equal to n 0 then what we can say after this is that if that is the case for example apply this for n equal to n 0 and you will get I will continue there so if you apply it for n equal to n 0 you will get the following that is a n 0 plus 1 this will be less than or equal to beta times a n 0 then a n 0 plus 2 will be less than or equal to beta times a n 0 plus 1 so it will be less than or equal to beta square times a n 0 so you can say for example a n 0 plus 2 this will be less than or equal to beta square times a n 0 and in general you can say that a n 0 plus k will be less than or equal to beta to the power k a n 0 in other words you can compare this given series to the series beta to the power n sigma beta to the power n and this is a convergent series not beta to the power n let us say beta to the power n into a n 0 so for n because on equal to n 0 each term of the given series sigma n is less than or equal to the corresponding term of this series and this is a convergent series and hence by the comparison test the given series converges so that is the argument in this case of course given a series of positive terms you can either apply root test or ratio test to decide the convergence or divergence of that series of course it can happen that neither the test is applicable in general the ratio test is easy to apply compared easier compared to the root test but root test is more applicable there are there are there are instances where root test will give some positive answer but the ratio test will fail one can easily construct examples like that but let us not go into that kind of thing since we are here at this convergence or divergence of the series let us also discuss one or two points which follow immediately from this let us also look at the series like this sigma n going from 0 to infinity a n x to the power n such a series is called power series of course this is again when we subsequently when we discuss the sequences and series of functions we shall come back to the series like this so this is a special case of this the reason for discussing it now is that certain things about the power series follow immediately from whatever we have discussed so far and so I am discussing that here or more generally we can think of a series like this a n x minus a to the power n all right now this is called power series with center at a and whenever we are given a series like this the question to be asked is for what values of x does this series converge and obviously for what values of x does it diverge those are the questions you can notice one thing very clearly here is that you can make a change of variable you can write let us say y equal to x minus a we can write then this series becomes a n y to the power n and so whenever the series converges for a particular value of y it will converge for the that that value of x you can put y equal to x minus a and draw the appropriate conclusions for x in other words what I want to say is that it is enough to discuss it is enough to discuss the series like this sigma a n x to the power n n going from 0 to infinity which is same as saying that without loss of generality we can assume that the center is 0 and it is enough to discuss that case and if center is something else we can easily draw the conclusions about those series by using the knowledge about this series now if we instead of this series now suppose I look at this is sigma n going from 0 to n mod a n x to the power n then we know that whenever this series converges then this is also because we have shown that every absolutely convergent series is convergent so we look at this and let us apply for example the root test for this so if you look at the nth term here it is a n x to the power n so if you look at the nth root of that nth root of mod a n into mod x to the power n then that is nothing but nth root of mod a n into mod x so if you look at the limb soup of the whole thing that is nothing but this is independent of n so it will be limb soup of this multiplied by mod x so that is why what is done usually is that what is done usually is that we take let us say l as limb soup of nth root of mod a n and what it means is that if l into mod x is less than 1 if l into mod x is less than 1 then this series converges if l into mod x is bigger than 1 the series does not converge and if l into mod x is equal to 1 we can cannot conclude anything from this information and since every time we are talking of 3 l into mod x is less than 1 etcetera it is convenient to take this number r as 1 by l r as 1 by l and then say that and say that whenever mod x is less than r that is same as saying that l into mod x is less than 1 so what we can say from here is that if mod x is less than r then this series converges then and which is same as saying that this series sigma a n x to the power n converges absolutely and similarly one can write if mod x bigger than r mod x bigger than n then we can similarly show that this series diverges and we have to discuss if mod x equal to r we do not know we have to we have to test that case independently. Now what we can see from here is that if mod x is less than r that is same as saying that x lies between minus r to plus r that is same as x lies between so that is if you look at this interval minus r to plus r so what we are saying is that whenever x is inside this interval the series converges absolutely and whenever x is outside this interval the series diverges and the end points of course we do not know we have to test it independently and that is why this interval is called interval of convergence of the power series we call this interval of convergence interval of convergence and this r is called the radius of convergence. Let us take one example this one example which we have already seen sigma x to the power n n going from 0 to infinity this is the geometric series and we have seen that this series converges if mod x is less than 1 and diverges in this particular case it will diverge for mod x because that is it diverges for mod x equal to 1 and it also diverges for this end points also r is plus 1 and r is minus 1 now what is to be noted under this is that we are here we are using the extended real length that is if this l is infinity then the radius is 0. We make that convergence if r is equal to 1 by l with the convention that if this l is 0 r is infinity and if l is infinity r is 0. So, to get an example like that let us look at this next example sigma x to the power n by factorial n n going from 0 to infinity in this case it is easy to look at the ratio test instead of the root test because here a n is 1 by n factorial. So, if you look at a n plus 1 by a n it is 1 by n plus 1 factorial divided by again 1 by. So, it is n factorial by n plus 1 factorial. So, it is 1 by n so 1 by n plus 1 right 1 by n plus 1 and that tends to 0 as n tends to infinity and using this we can show that that is same as saying that this of course I should have taken n through of mod n, but we can show that that is also 0 n through of mod n that is also 0 and then that will that will say that l is 0 that is saying that r is infinity that means this series converges absolutely for every value of x. So, this converges absolutely converges absolutely for all x in r this is what you already called this is what is called geometric series you perhaps also know that this is called exponential series is this whatever we have done so far. So, now let me take a special case of this I take x equal to 1 take x equal to 1 then this will be sigma 1 by n factorial n going from 0 to infinity and our idea is to show that the sum of this series is e this is what we want to show what is it that we already shown about this number e it is the limit of this that is suppose you take x n as 1 plus 1 by n raise to n then we have shown that this tends to e and now what we want to show is that limit of this is e limit of this is e means we already know that this converges from whatever we have discussed we already know that this series converges it converges means what it sequence of its partial sum converges. So, suppose we take this sequence it is s n is this 1 plus 1 by 1 factorial it is starting from 0. So, it is etcetera up to 1 say 1 by 2 factorial up to 1 by n factorial 1 by n factorial and suppose since we know that it is converges suppose we call that limit of this as let me just say suppose s n tends to s as n tends to infinity then that is then what we want to show is that s is equal to e what we want to show is that s is equal to e now in this case let me also mention see essentially what we are saying that the number e can be defined in more than one way this is one way you get the number e this is another way which you get number e in fact it can be also defined in one or two more ways only thing is that whenever you do it you have to show that all those definitions are the same. So, since we are not going to all those things let me just give you one reference it is a book by one K. D. Joshi is a professor in IIT Bombay K. D. Joshi the title is calculus for scientists and engineers calculus for scientists and engineers in this book there is a section and the title of that section is this e equal to e equal to e and you can predict from this title that the number e is defined in three different ways and this section shows the equivalence of all those three definitions. So, two of those ways are these what we are there is one more way. So, if you are interested in that you can just have a look at this for the time being we shall just show the equivalence of these two definitions. Let me recall that while when we are when we prove that this sequence converges we had got this equation xn is 1 plus 1 by n raise to n and then we had expanded this by using binomial theorem. So, it is 1 plus 1 plus we had it at n into n minus 1 by 2 into 1 by n into n minus 1 into 1 by n square etcetera and then we had simplified this to the following 1 plus 1 plus 1 into 1 minus 1 by n into 1 by 2 factorial and similarly the next term will be 1 into 1 minus 1 by n into 1 minus 2 by n into 1 by factorial 3 etcetera. So, suppose we go on like this the m th term will be 1 into 1 minus 1 by n into 1 minus 2 by n etcetera 1 1 minus m by n into 1 by m factorial this way and at that time we had noticed that each of these term 1 minus 1 by n whatever factors are occurring here they are all less nor equal to 1 you are subtracting some positive quantity from this. So, this will be less nor equal to 1 plus 1 plus 1 by factorial 2 etcetera up to 1 by n factorial and which is nothing but Sn. So, suppose you recall this what does this prove that xn is less nor equal to Sn for all n xn is less than or equal to Sn for all n. So, taking the limit what follows so limit of xn should be less than or equal to limit of Sn. So, we have called this limit of this as S limit of this as E. So, this proves that E is so this implies that E is less than or equal to S our aim is to show that E equal to S remember this is what we want to show to show E is equal to S and now what we are shown is that E is less than or equal to S. So, what is left which is also show that S is less than or equal to E is also show that S is less than or equal to E. Now, from this calculation you will see that suppose I take some m which is less than n suppose I take some m which is less than let us say 1 less than or equal to m less than n. Then what we can see from here is that xn is this going up to this nth term. So, suppose I ignore all these subsequent terms then I can say that xn is bigger than or equal to this sum up to this. So, what I can say is that xn is bigger than or equal to 1 plus 1 plus 1 minus 1 by n into 1 by factorial 2 etcetera. Let us say let me write one more term plus 1 into 1 minus 1 by n into 1 minus 2 by n into 1 by factorial 3. Suppose I go up to m and stop there. So, that will be 1 into 1 minus 1 by n into 1 minus 2 by n etcetera. It will be I think this should have been m minus 1 by n here because you compare with the previous term it is 2 by n and then followed by 1 by factorial 3. So, this goes up to 1 minus m minus 1 by n into 1 by factorial m. Now, this is true for all n satisfying this for you can say that this is true for all n bigger than m. So, what I say now is that keep some value of m fixed. So, keep m fixed and let m vary keep m fixed and let m vary let n be bigger than n. So, keeping m fixed and letting n vary suppose I take the limit of both sides suppose I take the limit of both sides then what is the limit of this? This will go to e this will go to e what will happen to the left hand side? Remember m is fixed. So, this term 1 by n will go to 0. So, similarly 1 by n 2 by n etcetera m minus 1 by n all of them will go to 0. So, that means these terms are going to become 1. So, what we can say is that that will imply that taking limit e will be bigger than or equal to 1 plus 1 plus 1 by factorial 2 plus 1 by factorial 3 etcetera up to 1 by m factorial that is the thing, but s m that is the thing, but s m and what did we prove now? We proved that e is bigger than or equal to s m, but after taking the limit with respect to n e is bigger than or equal to s m we can say for all m e is bigger than or equal to s m for all m and hence we can say that since e is bigger than or equal to s m for all m e is also bigger than or equal to s. So, this implies e is bigger than or equal to s. So, what it shows is that these two limits e is equal to s are the same that means the number e can be defined in these two different fashion. It is a limit of 1 plus 1 by n raise to n and it is also a limit of this series. Suppose now we want to also say how this s n this is a partial sum of the series and so that means s n goes to e means for large values of n the difference between s n and e is small that means you can take s n as an approximation of e s n as an approximation of e for large values of n. Now the question is suppose we want to know how good or bad this approximation is for example I can have a problem like this I want to find a value of e let us say correct to two decimal places. Then I should know what is the difference between then I should know what n I should take so that difference between e and s n is less than 1 by 100 that will be correct to then I can take s n to be the value of e correct to two decimal places. So to do that we should have some estimate for the difference between e and s n of course one thing is there all of these s n are less than or equal to e. This is a series of positive terms and s n is a monotonically increasing sequence that is converging to e. So this e is the least upper bound of s n so this is something that we know already that s n is less than or equal to e or which is same as saying that 0 is less than or equal to e minus s n. Now let us have some estimate for e minus s n so what is e minus s n s n is e is this let us take this definition of e is this and s n is up to 1 by n factorial. So what can we say about e minus s n e minus s n should be the remaining terms of this series which is sometimes called a tail of the series. So let us have some estimate for that so e minus s n will be this 1 by n plus 1 factorial plus 1 by n plus 2 factorial etcetera plus 1 by n plus 3 factorial etcetera. Now suppose I want to have some estimate for this term then what I can do is that I will take this 1 by n plus 1 factorial as a common factor from everything and write the terms which are inside as 1 by this will be 1 by n plus 2 this will be 1 by n plus 2 into n plus 3. Now there is one observation here this term is one I do not do anything about this is it clear that this is less than or equal to 1 by n plus 1 is it clear that this is less than or equal to 1 by n plus 1 square and subsequently subsequent term will be less than or equal to 1 by n plus 1 cube is that clear. So we can say this is less than or equal to this 1 by n plus 1 factorial into 1 plus 1 by n plus 1 plus 1 by n plus 1 square etcetera. Now what can you say about this last thing what is occurring in this bracket is a geometric series it is a geometric series and its common ratio is 1 by n plus 1. So we can take the exact sum of this series and what is that sum it is this 1 by n plus 1 factorial it remains as it is and this will be 1 minus 1 minus 1 by n plus 1. So what we will get after simplification it will be n plus 1 by n. So here you have n plus 1 factorial so that 1 by that 1 n plus 1 will cancel. So what will remain is this 1 by n factorial into n 1 by n factorial into n. So the difference between e minus sn e and sn will be always less than or equal to 1 by n factorial into n and this tells you how many terms you should take to get the good approximation of e. For example, suppose this number is less than say 1 by 100 if this number is less than 1 by 100 then it means that that sn is the correct approximation of e up to 2 decibels less. And what is that for example this number is going to grow very going to become the n into n factorial is going to become very large for even for small values of n. For example, if n is equal to 3 factorial is 6 so it is 3 into 3 factorial that will be already 18 if you take it is n as 4 it is 24 into 4. So it is already bigger than 100 so it will less than 1 by 100. So you can easily calculate the values of e value of value various values of e correct to whatever approximation you want. Now this also has another very interesting consequences and that is not very easy to prove otherwise and namely that let us let me write that as a theorem e is irrational by the way what we have proved here is this let me write it here. So what we have put is 0 less than or equal to e minus sn and this is less than or equal to 1 by n factorial into n for all n this is important till now you only know the how to prove that root 2 is irrational no other number you know how to prove that it is irrational. Now let us take this proof it is it is quite different from the proof that root 2 is irrational and for proving that we shall use this anyway how does one go on proving that something is irrational. Suppose e is irrational means what e must be of the form p by q suppose e is of the form p by q where p and q are integers and q not equal to 0 can we also assume that p and q are positive because we already know e lies between 2 and that is something we know. So p and q need not be we need not take negative number so in fact I can say that instead of this I can simply say p and q belong to n that is clear alright. Now look at this this is true for every n this is true for every n so I can take n is equal to q I can take n is equal to q. So what I can say is that look at I will rewrite this in a slightly different manner I can say this from here it also follows that 0 is less than or equal to n factorial into e minus s n this is less than or equal to 1 by n for all that is also true right. I am just multiplying by n factorial to whole t and so suppose I use that for n factorial then what it will mean it will mean that 0 is less than or equal to n factorial into not n factorial q factorial into e minus s n e minus s n is e is p by q minus s minus s q and less than or equal to 1 by n sorry 1 by q. Now just look at this can something like this happen what can you say about this number here see let us look at this q factorial multiplied by p by q should that be an integer because it is this will be q minus 1 multiplied by p all right. Then s q multiplied by q factorial should that be an integer because what is s q remember what is s q s q is 1 plus 1 plus 1 by factorial 2 etcetera up to 1 by factorial q. When you multiply that by q factorial is it clear that what you are going to get is an integer in fact a positive integer. So this number q factorial into p by q minus s q this must be a member of n this must be a member of n all right. And so what does it mean that there exist some natural number between 0 and 1 by q is that possible this is no natural number between 0 and 1 by q that is not possible. So that is a contradiction this is a contradiction it is a contradiction to what contradiction to this assumption that e is rational we assume that e is that is how we got we started with this e is equal to p by q and so that true that e must be rational. Now let us now come back so far we have discussed the cases when the series consisted of non negative terms and all the consequences of taking the non negative terms. Now how does one discuss the convergence or divergence of a series when the terms are not necessarily positive of course one thing is that we can discuss absolute convergence if you take the absolute values then all the terms becomes non negative and then we can apply. But that only tells you about whether a series is absolutely convergent or not. But it can happen that the series converges but it is not absolutely convergent and how does one go about testing those things of course the actual fact is there are very few methods of testing that kind of convergence and we shall discuss only one such case and that is the case what is known as Abel's test and Abel's test is applied to what are known as alternating series. Let me again repeat there is there are no good tests for testing the convergence of totally arbitrary series of positive and negative considering both positive and negative terms. So, this is one of those test alternating series means what the series changes sign alternately. So, the convenient way of representing that is you assume that all a i's are positive all a i's are positive let us let us say or let us say a n because 0 for all n and then assume that the series is like this a 1 minus a 2 plus a 3 minus a 4 etcetera or which is same as series is not sigma a n but it is sigma minus 1 to the power since we are starting a 1. So, minus 1 to the power n minus 1 into a n going from 1 to n this is the series that is called alternating series and what does Abel's test says that suppose you have alternating series like this that means the terms of the series are like this minus 1 to the power n minus 1 that is the series changes the sign alternately and suppose the following thing happens. Suppose this a n's are modulated decreasing suppose a 1 bigger not equal to a 2 etcetera in general a n is bigger not equal to a n plus 1 of course remember that this a n's are not a terms of the series terms of the series are plus or minus a depending on what is the value of n and this is one requirement and limit a n is 0 limit a n n tends to infinity is 0 then this series converges not sigma n this series converges then converges and we have seen that one of the ways of showing that a series converges is that its sequence of partial sums converges in fact that is the only way really speaking if you do not know any other tests etcetera and to show that a sequence converges one of the ways of showing that it is a Cauchy sequence and how does one show that a sequence is Cauchy that again the usually that is so you should show that for the large values of n and m difference between s n and s m can be made small. So suppose we take something like this that is 1 less not equal to m less than n and look at s n minus s m s n minus s m now s n minus s m will be what it will be the corresponding terms of the series starting from m plus 1 m plus 2 etcetera now only thing is we do not know whether m that m plus 1 is positive or negative, but that depends on whether m is even or odd let us to begin with let us assume that m is even so suppose suppose m is even suppose m is even then s n minus s m will be the first term will be a m plus 1 then minus a m plus 2 etcetera then again plus a m plus 3 etcetera it will go up to let us say plus or minus n I do not bother whether it is plus or I will come to that little later. Now what we can say from here is that this will be less than or equal to this will be equal to a m plus 1 minus this a m plus 2 minus a m plus 3 and then minus next a m plus 4 minus a m plus 5 etcetera now if the last term is depends it will be it may be either minus a n minus 1 minus n or it may be just minus a n depending on whether n is even or odd in fact that does not matter what is important here is that each of these terms are non-negative that follows because of it is it is non-negative decreasing so each of these terms are non-negative so that means you are subtracting some non-negative quantities from a m plus 1 which is positive a m plus 1 so does it follow from that this means that mod s n minus a m sorry mod s n minus s m this is less than or equal to a m plus 1 this less than or equal to a m plus 1 only problem is that we have only taken the case when m is even we have only taken the case when m is even if m is odd what will happen this will be minus a m plus 1 n plus a m plus 2 etcetera now how to deal with that case what I will say that after all you want mod s n minus s m you look at s m minus s n then that will be again a m plus 1 minus a m plus 2 etcetera and again you can do the same thing right so whichever way you take it really speaking it does not matter so even if the what I want to say is that you will always show that plus or minus of this quantity will be less than or equal to a m plus 1 in all cases so whether it whether m is even or all this will be the case is that clear so this means mod s n minus s m is less than or equal to a m plus 1 always now what can we say about this quantity a m plus 1 you look at this last condition here we have assumed that limit of nth term goes to 0 nth term goes to 0 so this tends to 0 as m tends to infinity but we assume that m is less than n so if m and n both tend to infinity which means you can given any epsilon we can always find some n 0 large enough such that wherever m and n both are bigger than that n 0 this a m plus 1 is less than epsilon but in that case mod s n minus s m will be less than epsilon that is same as saying that this is a Cauchy sequence and that is same as saying that the series converges is that clear so that is the test for convergence of alternating series and one of the well known examples of this where this is applicable is this series 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 etcetera you can see here we have taken a n as 1 by n a n as 1 by n so this is how auto decreasing sequence it converges to 0 and that is why this series converges this series converges to at the same time it does not converge absolutely because you take absolute values it will be 1 plus 1 by 2 plus 1 it will be sigma 1 by n and we already shown that that series diverges so this is an example of a series which converges but does not converge absolutely I think we will stop with that we shall make a few observations about this series of this type and then we shall proceed to the next topic in the next class.