In this video, I go over an infinite series in which arises an example of the famous telescoping sum such that all the terms of the series cancel except for the first and last term. The series with...
Transcript
Example 6, which states show that the following series is convergent and find its sum. So we have summation from n equals 1 to infinity or sigma, and then of 1 divided by n times n plus 1 here. So like that, it's very interesting here. So solution, this is not a geometric series, yet it's not written in that format we're used to of the A, R, n minus 1 like this. Yeah, it's not written this format like that. So we're dividing by the terms. So it's not a geometric series, so we go back to the definition of a convergent series and compute the partial sums and take the limit. So what we'll do here, just look at a partial sum here. S-N is equal to, so yeah, we can't just use a formula A divided by 1 minus R anymore. So we'll just use n, I equals to 1 of... Yeah, just use the partial one, so we'll replace N with I. 1 over I, I plus 1, like that. And yeah, let's just write out the full sequence or series out here. So we have, for I equals 1, we have 1 over 1 times 1, yeah, this is 1 plus 1, that's 2. And then plus 1 over when I is 2, we have 2, we have 2 times 2 plus 1 is 3. So 2 times 3, plus, and then we have a 1 here, 3 times 4, and just keeps going on and on. Then plus all the way here, where until we have i is equal to n. So have n times n plus 1. like that. All right, so we could actually simplify this expression even more, so instead of dealing it with like this, yeah, dealing with this 1 over i divided by i plus 1, we can separate this up into partial fraction, so we can simplify this expression if we use the partial fraction decomposition, see my earlier videos for that, or just use partial fraction decomposition. So you can see my earlier video there is link to it. So what we'll do here is, well, let's let, let's let, this term right there, that sequence, i, 1 over i, i plus 1 equals 2, and break it up into into its fractions so a over i plus b over and then we have i plus 1 like that and now what we have to do is just find the coefficients a and b and then seeing what they are then we could basically simplify this this term into these simpler terms like that so this equals to or what we'll do is multiply the top and bottom to get this common denominator. Yeah, so thus the first term we'll have A, so we'll have here is going to be A over I times it by I plus 1 over I plus 1, so that we have the same common denominator, we can add those up, plus, then we have here is, yeah, then we'll multiply this by I over I times by over here, B, i plus 1 like that so calculate this out we're going to have a i plus a plus b i like that all over all over i i plus 1 plus 1 like that and this left side is equal to just 1 over i i i plus 1 like that Yeah, so that's what we have is, well, the bottom part just cancel, so we're just left with, well, 1 equals to A, I, or it actually just factor out the I, so we'll have I like that, and then A plus B, and then plus A, like that. And now, since there's no I on the left side, well, this needs to go to zero. Yeah, so the only way the left side can equal to the right side is if this whole thing vanishes goes to zero, so that's what we have is A equals to negative B like that, or I'll just write A plus B equals here. So if this goes to zero, we have A equals to 1, so that this left side is equal to it. So A equals to 1, so then what we have is B equals to this over negative A, which equals to negative 1. So circle this, and circle this right here, this equals to B. So that's what we have is the simplified equation. This right here, 1 over I, divided by I plus 1. So we have 1 over I times I plus 1 equals 2, A is 1, B is negative 1. So that's it, so 1 over I minus 1 plus 1. So all we do is added minus there. And I'll just circle, this is pretty interesting. decomposition of like that. Yeah, so now that we've broken up like this, what we could do now is add these up, and we get a very, very fascinating result. So basically the same thing here where we're adding this up, but use it for this setup. So now what we have is partial sum as n is equal to, and then infinity or a summation like this from i equals to 1 up to n of 1 over i, i plus 1. So now this equals to the partial sum above there. I mean, of this partial decomposition, partial fractions. So we have n, i equals 1. Now we can just add these up like that. 1 over i, minus 1 over i plus 1. And now what we'll get is a very, very interesting result, and this is called the telescoping sum. So when we write these terms out, so for i is 1, so we have 1 over 1. That's right, 1 over 1, minus, then here is I plus 1, so 1 over 2. So we have that term. Now plus, then when I is 2, we have 1 over 2, and then minus, and then over here, 2 plus 1 is 3. So 1 over 3. And likewise, keep going on and on. You're going to go over here. Now we have for when I is 3. we have 1 over i is 1 over 3 minus and then we have 1 over 3 plus 1 is 1 over 4 and it keeps going on and on all the way until we have i is n so we have plus 1 over n minus 1 over n plus 1 like that but notice what we have here so we have 1 over 1 which is just 1 and then we have minus one half, but the next term we add one half. So this cancels with this. Now we have here minus 1 3, plus, and then right after 1 over 3 like that. So this would cancel. Likewise for 1 over 4, there's going to be another 1 over 4 over 4 over there. They're going to cancel all the way until 1 over n, then this 1 over end cancels like that, and we're just left with the first term and the last term. So amazing. So we just get is s n equals to 1 minus. 1 over n plus 1, and this is absolutely fascinating. And here, yeah, I'll just have to add another hashtag amazing for this. Yeah, these kind of derivations and this kind of stuff that arises in math is what got me interested in just learning more about mathematics. Yeah, learning more on my own, because it's pretty fascinating stuff. So, yes, notice of the terms cancel and pairs, every single one just cancels like that in the middle. This is an example of a telescoping sum, because of all the cancellations, the sum collapses, yeah, like a Pirates collapsing telescope, yeah, right here. into just two terms. It just collapses all into two terms, like a pirate collapsing telescope. Very, very interesting. So here's a screenshot out for this collapsing telescope right here. So yes, I actually have one like this small. So you can have it here and expands out, or you can compress it, to all you left is being in the end. Very fascinating. So thus what we have now is to get the limit of that sum. So we have a partial sum, so thus the limit as an approach is infinity. and approaches infinity of s n equals to the limit as n approaches infinity of the partial sum, which is over here 1 minus 1 over n plus 1, 1 like that. So here we have 1, then here 1 divided by n plus 1, this n just goes to infinity. So in other words, this is, we have a scenario that is like this, we have a 1 over infinity, in other words, this goes to 0. So we have equals to 1 minus 0 equals to 1. So very, very fastening. So that telescoping example, and this one goes to 1, the telescoping sum. Yeah, so therefore, the series or the given series is convergent, and for completeness, we have the summation, infinity, n equals to 1 of 1 over n, n plus 1, like that. This equals to 1. Yes, and going further on this example, the following figure below illustrates example 6 by showing the graphs of the sequence of terms A -N equals 1 over n times n plus 1, and the sequence SN of partial sums. So notice that the sequence A-N approach 0 as S-N approaches 1. So there is the partial sum, the total sum, is approaching 1 here, so that's 1. But then this one's approaching zero, but the sequences or the terms themselves are approaching zero, as I also illustrated in earlier examples. It's very, very fascinating. So at the end of this video, I go over two exercises on two geometric interpretations of example six, which are quite fascinating, so stay tuned for that.