 I think last class we were talking about the convergence test right and I think I stopped at the limit comparison test correct yeah so we covered the nth term test we covered the p-series test we covered the comparison test we covered the integral test so now we are going to start with the ratio test so let's start with the ratio test the ratio test okay again in the ratio test let's say you have been given that summation an is a non-negative series this is a non-negative series I've already explained what's the meaning of a non-negative series a series where all the terms are positive yeah okay let the limit of n plus 1th term to the nth term be equal to L okay now this says that if your L is less than 1 the given series converges okay so this L is a finite number and if this number is less than 1 the series converges and if L is greater than 1 the series diverges so equal to 1 okay and equal to 1 the test fails so that test fails that means it is inconclusive okay now many a times it's a good practice to take the mod of this term normally we talk about mod of an plus 1 by n as n tends to infinity and I find this value L okay but since we have already told that the series is non-negative we actually don't need this mod okay but many a times you will see that they'll use the mod also sometimes doesn't make much of a difference to the test so remember the ratio test you take the n plus 1th term to the nth term ratio and find the limit of it as n tends to infinity if the limit which is a finite quantity L if that is less than 1 it is a convergent series if it is greater than 1 it's a divergent series but the series but the test doesn't say anything if your L becomes equal to 1 okay let's take few questions based on this does summation 1 by n factorial converge or diverge so if I have to find this by using our the ratio test what I'll do is I'll write the a n plus 1 at term which is 1 by n plus 1 factorial a n is 1 by n factorial and now I'll find the limit of a n plus 1 by a n that is 1 by n plus 1 factorial divided by 1 fact 1 by n factorial will be like this that's nothing but n plus 1 and as n becomes very very large this term will become 0 so since zero is less than 1 remember it would be a convergent series it would be a convergent series okay let's find another one does summation n to the power n by n factorial converge or diverge try this out so what is a n plus 1 n plus 1 to the power n plus 1 by n plus 1 factorial right okay a n is n to the power n by n factorial correct yeah so let's find out the limit as n tends to infinity of a n plus 1 by n okay and that would be limit n tending to infinity n plus 1 by n plus 1 factorial by n to the power n and we'll have n factorial here as well okay so if I'm not wrong it gives you a limit n tending to infinity this will cancel with an n plus 1 over here yeah so one of the n plus 1 will go off so it'll give you n plus 1 by n to the power of n correct which is nothing but 1 plus 1 by n to the power of n now this type of limit is actually termed as 1 to the power infinity form of limit and I think I've done a similar type in the class as well yeah so this answer is going to be now in this case the answer for this is e ready to show how it is e yeah you would have done this in your ncrt also some basic limits are mentioned one of them is this okay why is it e why it is e okay so in this case what do we do is we'll consider this limit to be log to be l sorry and we take a log on both the sides so let's take log to the base e on both the sides log to the base e okay so this becomes limit and tending to infinity and log 1 plus 1 by n what do we do is we take 1 by n as let's say y so this term becomes limit y tending to zero this will become ln 1 plus y by can I write this can I write this n as this divided by 1 by n okay and this entire term is log of l this is a limit which you have already seen before aditya yeah ln 1 plus x by x x tending to zero that's 1 so log l to the base e is 1 so l becomes okay yeah now e as we all know so it diverges right like e as we all know is 2.71828 da da da da da which is greater than 1 and hence this particular series will diverge so it's a divergent series so what he said is of 1 to the power infinity form so but that doesn't give you e right like why not who told you doesn't 1 to the power infinity means this one is not exactly 1 it's tending to 1 again 0 by 0 is not exactly 0 by 0 it's tending to 0 by tending to 0 yeah right yeah so with this we come to the end of our series test now we'll be starting with alternating series so did you say anything my internet connection when did you say anything before this I said that we have ended the test on convergence now we are going to start with alternating series now what's in alternating series alternating series is basically where you have alternating plus minus signs between the terms okay something like this a1 minus a1 minus a2 plus a3 minus a4 etc ok let's say it goes to infinity and there in between you have a term minus 1 to the power of k plus 1 a k that this is called an alternating series you can abbreviate this series like minus 1 to the power k plus 1 a k k going from 1 to infinity remember we normally choose our a k to be positive so the sign are external to it and all these terms themselves are positive terms okay one simple example of this would be a 1 minus half plus one third minus one fourth and so on okay like this okay we're intense to infinity this is called an alternating harmonic series this is an alternating harmonic series okay so we are going to take a test for these kind of series as well which we call as alternating series test so listen to this test very very carefully the alternating series test this alternating series says that an alternating series converges an alternating series converges if it meets the following two criteria number one it's a n plus one term should be lesser than a nth term for all n remember I'm still repeating this a k is taken as positive for all k okay so don't think that okay they have to be one of them have to be greater than the other they both are positive by the way we are putting external signs separately so remember I've written over here categorically that all terms are considered to be positive over here okay and second is the limit of the nth term as n tends to infinity should be 0 is that fine okay so let me ask this as a question to you does the series minus 1 to the power of n plus 1 by n we just now saw the series was called as the alternating harmonic series right does this converge does this converge or not converge or diverge answer this use the two tests use these two criteria mentioned the first one it passes me so a n will be what a n will be 1 by n right yeah a n plus 1 is less than this that's obvious so since or for the region for the reason a n plus 1 is less than 1 by n it passes the first test so it converges and not only that the limit of a n which is 1 by n as n tends to infinity is 0 so both you can yeah so it is a convergent series right and it's quite surprising because the harmonic series was not converging right but the moment you make the science alternating between the terms we get a convergent series okay now there's something called absolute convergence and conditional convergence let's look into this absolute convergence and conditional convergence what's an absolute convergence a series with a mixed sign or an alternating series is set to converge absolutely if the series obtained by taking the absolute values of the term converges correct see let's say you have been given a series like this okay get it a point so yeah if if you're if you take more you lose all the science that means if you're this converges then even this will be convergent and those type of convergence would be called an absolute convergence I'll give an example I'll give you a very simple example to understand this see we have already seen that oh so we have seen that 1 by 1 square 1 by 2 square 1 by 3 square da da da da da till 1 by n square this is convergent by the way if you remember it was actually 1 by n to the power p correct and p is greater than 1 so this is a p-series correct p-series is known to be convergent when p is greater than 1 correct so even if you take the alternating sign between the terms let's say like this that means minus 1 to the power let's say n plus 1 by n square da da da da da da this will also be convergent and such convergence would be called an absolute convergence are you getting this point evaluated say the one we wrote for the alternating harmonic series the sign didn't matter to us like at all even when we solved it I didn't get you like when we solve when we use the alternate series test for the alternating harmonic series the sign we didn't even the sign didn't matter at all in how we solved it no that that convergent is different now that convergent itself can be of two types it can be absolute it can be conditional so I'm not I'm not questioning its convergence if this test is satisfied it is convergent now the question may ask you what type of convergence it is is it an absolute convergence or is it a conditional convergence so all you need to check for convergence of an alternating series is whether these two criteria are met okay now in addition to it being convergent you're trying to identify the type of convergence it can be absolute it can be conditional so when do you call it as an absolute convergence it is called an absolute convergence even if you take the modulus of all the terms in that series and that series is also convergent then that alternating series would be called as absolutely converging series yeah but but try to take an example like this one by one one by two one by three etc which is a harmonic series right yeah we know from our experience that this diverges right but just now I took an example just now I took an example where I showed you where was that example yeah this example and we concluded that it was convergent correct so now this convergent would be called conditional convergence right so this diverges but this guy converges okay this guy converges and this converges convergence will be called as a conditional convergence just for your information because the question can directly be targeted at type of convergence also okay can we take some question yeah but there's no test for alternating test series test for divergence it's just that if these two conditions are not met it's convergence yes yes any one of these two condition is not met it is it is divided let's take questions or determine whether determine whether determine whether this particular series converges absolutely or converges conditionally or actually diverges which are the three options yeah absolutely it diverges because as you would see the limit of a n this one as n tends to infinity is actually as you rightly pointed out it's one if you find this limit as n tends to infinity we need to divide by n square I'm trying to copy Sal Khan of Khan Academy so it will be it will be one right I some feel that he puts on that lot of accent sometimes it makes very difficult to understand what he's saying but he was raised in US oh he's actually a Pakistani right yeah now Bangladesh yeah okay and Sal Khan next question is did the mind whether that the mind whether this is cube root of n plus one okay and equal to one to infinity the mind whether this converges absolutely converges conditionally or diverges does this converge absolutely does this converge conditionally or does it diverge what do you think share so I think when you say it converges absolutely that means you are trying to claim that this particular series converges right yeah you really think so see try to compare this with one by try to compare this with the p series okay now in this case you realize your P is less than one yeah so it diverges right yeah so if it diverges that means it cannot it cannot converge you can this my required series will never convert absolutely it can either converge conditionally or it can itself diverge only two possibilities are there either first one or second one okay so first one is rule out now let's check the alternating set test series condition that is is a n plus one less than n that means is n plus one plus one cube root less than one by n plus one cube root yes this is correct this is true this is true because the base here is heavy correct and what about the limit and tending to infinity of one by n plus one whole cube absolutely this is going to be zero so yes it is converging okay it is a convergent series but because the positive counterpart is divergent we will call it as conditionally convergent so but this is one by n plus one to the power p right so we just say that because one by n to the power p is divergent and that's a lesser function no again see what did I say if you start putting n value as one you get the first term like this second term like this third term like this like so on yeah now do you recall the theorem where I said that even if you exclude or include certain terms it doesn't change the nature of the series okay so you exclude the first term and right so we know that this is divergent and if you remove this also if you remove this term also the remaining series will still remain a divergent one doesn't make any difference even if you remove five six ten or finite number of terms it will still remain divergent it's like you're pulling out few drops from an ocean doesn't doesn't reduce the ocean by you know a marginal quantity yes or no okay so the answer here is this particular series is conditionally convergent next important thing that we are going to talk about today is approximating the limit of an alternating series approximating the limit of an alternating series okay now I try to understand this through a simple example because does the theory is very confusing in this okay see when you're trying to sum up alternating series let's say I give you a series like this summation of minus 1 to the power of k plus 1 by k k from 1 to infinity so we know that it's an alternating harmonic series correct okay it goes like one half one third minus one fourth plus one fifth minus one sixth and so on correct let's say I call this series as s okay now if you try to approximate these series by taking a certain number of terms let's say I take the first six terms okay normally we call this as s6 okay now the theorem says that if you want to approximate an alternating series by taking s6 remember that there will always be an error in this which would be actually lesser than the first omitted term modulus which we call as r6 r6 is basically nothing but the first omitted term in that series omitted term means since I considered s6 ideally if you have if I would have continued I would have got 1 by 7 right this is the first omitted term in that 7th term that is omitted because we were asked to only take s6 okay okay okay okay okay so just called r6 but it's basically the seventh actually it is actually this is called r6 and this r6 is actually lesser than that we say mod of this is lesser than that omitted term which in this case is your 1 by 7 why am I raising this yeah 1 by 7 are you getting this fact yes okay I'll try to explain this with a simple example in this case itself if you're asked where what is the limit of this sum that means if you ask to enclose that sum within two boundaries okay and you're only allowed to take the first six terms in your account while trying to estimate the value of s okay so think as if you want to estimate the interval in which s lies in which s lies only by using the first six terms okay I'm taking this a simple example in this case okay so then this example says that your s will actually lie between s6 and s6 plus mod of the omitted term where n here is 6 so that will be your seventh term I'll make this n as 6 are you getting this point so can you all use your calculator and tell me the value of this the first six terms even I can use that let me just use myself so 1 minus a half plus one third minus one fourth plus one fifth minus one sixth okay so this the sum s6 comes out to be point six one six okay so this series will approximately lie between point six one six and but will be lesser than point six one six plus one by seven that's point seven five nine are you getting this so this harmonic series will somewhat lie in this in this range are you getting this point now you would ask so if I would have taken you choose the inter do you choose the number of terms they tell you they will tell you they will tell you they will tell you so let's say you would have claimed here that sir had you chosen seven terms then this range would have been different that range would then be a subset of this that means you come more closer to your s right so let's say I take the seventh term let's say I want to approximate it between seven terms I would say s would lie between s7 and s7 plus more of the eighth term more of the eighth term so if I just include a plus seven let's see what happens minus one by seven okay if you do minus one by seven this would come out to be it was minus one by seven or plus one by seven it's so it's between zero point seven five nine and plus one by seven okay no point seven fine and it will be divided one divided by eight so it will say zero point seven five nine eight eight four and point eight eight four okay so more the more number of terms you take the more closer you will be to your final answer okay let's take questions on this so basically what I'm trying to say from the previous slide that the error the error that you get will always be less than the omitted term yeah are you getting it it's very important let's say question that can be framed on this using the alternative alternating series error bound use the alternating series error bound to determine how many terms must be summed to approximate to three decimal places the value of this series read the question once again use the alternating series error series error bound to determine how many how many terms must be summed must be summed to approximate to three decimal places to approximate to three decimal places the value of this s now what are the meaning of that the meaning is let's say I summed it till x number of terms or let's say m number of terms correct okay and I got some value to let's say a m I summed it till a m terms okay and let me use the sign as as well that's one and I got some value sm okay now which term is omitted a m plus 1 a term is omitted if the mod of this term in fact we already always take it positive so there's no point writing a mod because sign is always external so this is I'm what I'm calling as a k so if a m itself has a value where the first three decimal places they're asking you to approximated three decimal places right mm-hmm that means the first three decimal places don't change that means it should have a number appearing here it should have a zero zero zero appearing over it but is it necessarily out of the state can be any number approximate to three decimal places means whatever has happened over till sm the error here which is your a m plus 1 it should not change the first three significant figures here that means it should not take change the first three decimal places over here then only your s will be correct to three decimal places isn't it is that clear yeah okay so you tell me after which term your a m plus one will start becoming point zero zero something so lowest of this I'll take it as one so when is this lesser than this that's what I'm trying to say so basically you are trying to find out m plus 2 right this will become m plus 2 correct yeah when is m plus 2 sorry this is m plus 1 only because it's m plus 1 m plus 1 I'm sorry yeah when will this square become lesser than point zero zero zero one it has to be less than 100 see yeah I understood this why I'm taking three zeros let's say till sm let's say assume as you hypothetically let's say till the sum was point seven four three eight okay and this error this is actually your error bound the upper bound on the error is this okay so this error is zero point zero zero zero one and when you add this or subtract this it is not going to influence my first three decimal places right because when you add it or when you subtract it whatever you are doing it will remain point seven four three and this will change depending upon whether you are adding or subtracting at least leaving this term at least these three would be fixed so what if it's like zero point nine nine nine nine nine plus zero point zero zero one then it would go to then it would increase like number of places so we add this error actually right yeah so if you add zero point nine nine nine nine oh you're saying that if there's a nine over here yeah across all four of them like nine nine nine nine plus zero point zero zero one that would give you one right huh so wouldn't that be wrong that's actually a good question but I've never seen a scenario like that coming up whenever they give a question they assume that not all the series is nine okay okay not all the numbers is nine actually but like even if the last digit is nine it'll still make a difference but that matter yeah it may change this so ideally so ideally speaking yeah you would have but that would be like you know over being over cash over you know cautious with that particular series so you are you're taking some hypotheticals you're taking some extreme scenario and trying to be over cautious yeah it could be it could be so actually depends you should actually check but normally the value of M is so large that it is not literally possible to sit and you know add them in the examination all yeah so we tried to just take the you know optimal normal scenario rather than taking an extreme scenario yeah okay so basically M plus one square should be greater than 10,000 correct yeah yeah M plus one should be greater than 100 that means you should take more than 99 terms if you don't want your first treatment basis to change okay okay let's see whether if you have a question on this so the answer is just 99 terms yeah next question is if there is a series like this this is actually a series for tan inverse one which is actually pi by four so if you go all the way till infinity this value will approximately come out to be pi by four now the question is if this series is used to approximate pi by four with an error less than so your error should be less than 0.001 then the smallest number of terms needed is then the smallest number of terms needed is option A 100 option B 200 option C 300 option D 400 option E 500 so remember the maximum error possible Rn will always be less than An plus 1 this is the maximum error possible if you call as the remainder of the series that remainder of the series will always be less than the first of the omitted terms see actually it happens like this f1 f2 f3 let's say you stop at Sn okay right let's say you want to calculate this whatever is the remaining term we actually call it as Rn the remainder term this remainder term modulus will always be less than the first term in this particular remainder basically you stopped here right so after that it would have been like this had the series continued it would have been like this so this will be lesser than the first omitted term always this is what this theorem says yes and this is called the error box why does call an error is because you when you add this up and you subtract it I should not have written s1 s2 actually I should have written a1 a2 so when you write it as a1 minus a2 plus a3 and so on till whatever is your An okay so this is Sn so the difference between Sn Sn which is actually your Rn will always be less than the first omitted term that's what this series is so what is the general trend of the denominator if I write An term it's 1 by 2n plus 1 2n minus 1 yeah okay so let's say the omitted term is An plus 1 which is 1 by 2m plus 1 this error should be less than 0.001 so 2n plus 1 should be less than 1000 yes or no yeah so what are the answers I mean so 2n plus 1 is greater than 0.001 oh I'm sorry sorry sorry thanks so 2n should be greater than 999 by 2 and should be greater than 999 by 2 so next number after this is 500 so I have to at least take some till 500 so that the error in calculating the series is less than 0.001 so if you the the moral of the story is if you take 500 terms like this you can accept you can expect that the answer that you get is not far away from the original answer by more than 0.001 are you getting the moral of the story yeah this is the most confusing part of the the alternating series yeah you had written the An term is 1 upon 2n minus 1 but 1 will not satisfy that right it will be minus 1 why not so if you put n as 0 and n position of the term is always a natural number this is the solution you took it as 2n plus 1 because the error is n plus 1 the summing till this term but the error will be the first of the omitted term correct this is what I wrote your summing till n terms and the error whatever is left cannot be greater than the first omitted term so arc will always be less than a n plus 1 yeah understood yeah okay let's take another question let's take this question we have summation of this okay and this our series is approximated by the sum of first 300 terms this is you're trying to approximate it see this is a never ending series right so you want to approximate this sum and you try to approximate it by summing first 300 terms okay what is the max error that can happen in your by your approximation so what is the error bound of your approximation option a point zero zero one option B zero point zero zero two option C zero point zero zero five option D zero point zero one and option E zero point zero two so the error bound is which of the following I don't think so it should take so much of time this is your bound so I said so one nine hundred and three minus one it's one by nine hundred and three how much is this zero point zero one one one something so it's zero point zero zero one one zero point zero zero one right so let's talk about power series don't confuse power series with the P series P series is different power series is different so when you write a kind of infinitely long polynomial equation like this that means it's never ending okay goes to infinity then these kind of series are called as power series and this is every related as writing it as summation a k x to the power k and k is going from zero all the infinity so you can abbreviate this like this so whenever you come across such a term it basically should remind you of a power series now it's not necessary that the power series is always in terms of x you could have a power series like this as well so that's something like a zero a one x minus a given constant let's say a a 2 x minus a square a 3 x minus a cube and again it will go to infinity okay this is also a power series this is called power series in x this would be called as a power series in x minus a okay this very important for us to understand power series because there are two power series which we are going to deal with one is called Taylor series and others called the Maclaurin series we'll talk about it first try to understand so the previous one is it a 2 x cube or a 2 x slip of the pen now even for such series we need to find out the convergence and divergence that all depends upon what is your x what is the interval in which your x is lying so there is something called the radius of convergence now we'll talk about a very important concept which is related to power series convergence which we call the radius of convergence that means under what interval can your x lie so that this given series even if it goes to infinitely many terms is a convergent one remember here all the a is are constant a 0 a 1 a 2 etc they are all constants okay now in order to explain you the radius of convergence I will like to take a simple example so that it's clear to you what do I mean by radius of convergence let's say I give you a question find all values of x for which find all x for which 1 plus x plus x square all the way till x to the power n converges okay so this is also a kind of a power series only correct now do you remember the ratio test do you remember the ratio test yeah now here I will take the what are the ratio test can you recall from the infinity a n plus 1 by a n should be equal to and if l is less than 0 the lesson 1 sorry converges one diverges and equal to 1 you don't know very good it feels the test is correct now here since x involved as I told you mod will be very important for us because x can have positive negative sign so I don't know what is the nature of this because we don't know whether it's a positive series sorry whether it's a non-negative series or whether it's alternating series I don't know about it so it is suggested that when you are applying the ratio test to such terms where the nature of the terms themselves are not known to us clearly whether they are positive negative alternating it's advisable that you take a mod over it okay now simple question I would like to ask you what is the limit of n plus 1 it's term which is x to the power n plus 1 by x to the power n mod going to be you'll say simple is going to be mod of x yeah right because anyways you're going to lose all the n's and going to just get our x left okay now you're saying that if it is less than 1 only it will converge correct and you want it to converge so what will you say you'll say mod x should be less than 1 then yeah that means the interval of x that you desire for the series to be a convergent should be minus 1 to 1 this will be your radius of convergence or you can call it as the interval of convergence whatever you feel like okay so many books will say the radius is they'll they'll call one as the radius of convergence okay so normally it is defined as radius of convergence is defined as this value r r is called the radius of convergence so here one will be called as the radius of convergence and this would be called as the interval of convergence the moral of the story here is that you will be given a kind of a series and you need to just figure out if the series converges for this or alternately if your series diverges for this correct so if it converges for this value or you say this diverges for mod x greater than r then r will be called as the radius of convergence critical value r is named as the radius of convergence yeah okay so let's try to find out for other examples as well for what value of x for what value of x does this series converge again ratio test yeah so limit n tending to infinity mod of a n plus 1 by n that's going to be x to the power n by n plus 2 divided by x to the power n minus 1 into n plus 1 correct okay so this becomes a limit n tending to infinity and you can take n plus 1 by n plus 2 this thing and you can take a mod x okay so as n tends to infinity this limit is going to be 1 right so it's going to be giving you again mod x so if mod x is less than 1 according to the ratio test the given series would be convergent else it will be divergent okay is that fine okay now very important thing here is to note that this tells you that x should be between minus 1 to 1 right yeah it is not allowing you to take the value of 1 right now but if you see if you put x as 1 what do you get actually you get 0 1 let's say I put n as 1 what do you get you get a half minus 1 plus 1 fourth minus one fifth and so on this is what your series becomes isn't it yeah it's an alternating harmonic series and we have seen that it is conditionally convergent yeah so your answer would be slightly modified here instead of saying minus one to one you would say minus one to one including one this will become your answer do we always have to substitute the values and put it no in certain cases like this you need to just do a check at the corner points in certain cases where you have a feeling that this is going to become a harmonic series which is alternating then you should actually make this check okay and again with a lot of problems only you'll come to know but do I put a minus one in this no if I put a minus one in this I will get something like one third one fourth which is actually a harmonic series yeah but this is divergent so it is it is correct not to include minus one but one can be included because one will give you a case of an alternating harmonic series minus one the denominator will become zero oh x sorry x sorry let's see basically we know to include one yes be careful about such questions but there are some questions like the one which I'm going to give you right now let's say x to the power n by n factorial okay for what value of x for what values of x does this series converge so what it becomes like zero yeah absolutely yeah it's always converted yeah I guess absolutely it's convergent so if you do that ratio test limit n tending to infinity mod x to the power n plus 1 by n plus 1 factorial divided by xn n factorial this would lead to limit n tending to infinity x by n plus x by n plus 1 right which is actually a limit of n tending to infinity mod x by mod n plus 1 and no matter whatever finite quantity you have over here because of this heavy base over here it will become a zero and zero is less than one so this will be always convergent series next question for what find all x find all x that means find the interval of x for which the following series converges for which the following series converges the series is like this 1 plus x minus 2 by 2 factorial x minus 2 square by I'm sorry not 2 factorial is 2 to the power of 1 by 2 to the power of 2 x minus 2 whole cube by 2 to the power of 3 and so on till infinity so instead of giving you a general expression in summation form they are actually written in terms for you so that no they can check whether you are able to write the n nth term or n plus 1 th term like this what do you say so it converges right for what value of x does the following series converges minus 3 by 2 to 3 by 2 no this is 0 less than 0 no so minus 4 to 4 minus 4 to 4 is also not correct it's a less than minus okay it's minus see if you write the ratio of n plus minus term that doesn't make a lot of a difference because cancellation of terms will happen okay so that would lead lead leave you with what x minus 2 correct by 2 by 2 by 2 is less than 1 that means mod of x minus 2 should be less than 2 yeah that means x minus 2 should lie between minus 2 to 2 that means x should lie between 0 to 4 okay yeah that's what I meant so I just didn't do that thing is that fine okay so I think I would need one more class for to finish up the series sequence chapter because we haven't yet started with Maclaurin series Taylor series so one more class would be needed so we can have it I think tomorrow same time would you find would you find with you guys find it over and out from my side