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### Course: Calculus, all content (2017 edition)>Unit 7

Lesson 7: Series basics challenge

# Telescoping series

Telescoping series is a series where all terms cancel out except for the first and last one. This makes such series easy to analyze. In this video, we use partial fraction decomposition to find sum of telescoping series. Created by Sal Khan.

## Want to join the conversation?

• I remember that Sal's older video on partial fraction decomposition has a simpler way for solving for A and B:

After multiplying -2/[(n+1)(n+2)] and the fractions by (n+1)(n+2), we have
-2=A(n+2)+B(n+1)
and we solve for A by plugging in n=-1, and solve for B by plugging in n=-2.

Why does Sal solve using a system of equations?
(43 votes)
• I suspect that in this case he observed that the B term cancels if you add the equations, so doing it this way is pretty easy too. Plus, seeing a number of different ways to solve problems can be helpful.
(26 votes)
• I think the most important question here is how do we know if a given series will be telescopic or not? Do we have to resort to expanding the series for the first few terms? And, if we do that, how do we know that we aren't fooling ourselves into a self sense of security if it the first few terms work out but after perhaps the 100th term things go wrong?
(23 votes)
• When in doubt you would use an inductive proof. That is, demonstrate that it is correct for the first term (or enough terms to establish the pattern), and then demonstrate that if it's true for term n it has to be true for term n + 1.
(16 votes)
• Why is it called a telescoping series?
(5 votes)
• Have you ever seen an old-fashioned telescope made of two or more
tubes that slide inside one another, so you can compactly store it?
It's sort of like this:

o------------o#####o------------o
o------------o ######## o------------o|
o------------o##############o------------o |
| | <----- | |
o------------o##############o------------o |
o------------o ######## o------------o|
o------------o#####o------------o

A telescoping series similarly lets each term "slide" into the next,
"collapsing" the whole thing down to a very simple sum.

Perhaps that configuration is what prompts the name. It does look a
lot like a telescope or tripod leg being collapsed.
(6 votes)
• I forgot something. At the start of the video in the top left, what is that 'E' or backwards 3 looking symbol? Name and meaning/purpose?
(2 votes)
• It's a capital sigma, the greek letter Σ. It stands for summation. The "n = 2" below it tells that the variable n starts with the value 2, and the ∞ above tells that it runs to infinity.

For example:

Σ{n = 1 to 4} n
would mean
1 + 2 + 3 + 4
= 10

Σ{n = -1 to 4} n^2
would mean
(-1)^2 + 0^2 + 1^2 + 2^2 + 3^2 + 4^2
= 1 + 0 + 1 + 4 + 9 + 16
= 31

Σ{n = 0 to ∞} 1/(n + 1)
would mean
1/(0+1) + 1/(1+1) + 1/(2+1) + 1/(3+1) + ...
= 1/1 + 1/2 + 1/3 + 1/4 + ...
= ∞
(12 votes)
• Wouldn't it just be easier to expand the denominator and then take the limit as n goes to infinity?
(4 votes)
• Allie this was actually a really good question, and after reading it I was wondering the same thing. Eventually i came to the conclusion the reason it doesn't work is because, whilst this may work without the Summation (sigma) when taking a single term to infinite, it doesn't work when we're taking an infinite number of terms (even if they're very small) to infinite, because all of these terms are adding up and possibly converging to some value, in this case 2/3! That's how I understand it anyway.
(3 votes)
• Is there actually a formulaic way to recognize this solution/solve for it or do you just have to guess and look for terms to cancel out?
(3 votes)
• The first part, partial fraction decomposition is a typical way to handle this type of integrand. Actually usually you have a quadratic expression that you try to factor to get the integrand in this form in order to apply PFD). As for the telescoping part, consider this video as an introduction to that type of series - so now you are aware of them and can be on the lookout for them - when things seem to not be falling into place, check if the series is telescopic and apply a similar analysis.

More and more, the math of integration and the math that follows requires you to develop your intuition and creativity to come up with a solution. You will discover that many great proofs happened by a seemingly left field approach. To be able to do this demonstrates that you have a grasp on the implications, consequences and connectedness of the theorems and definitions you are learning.
(5 votes)
• What is the link to the video on partial fraction expansion?
(3 votes)
• I do not understand why the (n+1) is used instead of the (n+2)? They are the factors of (n^2+3n+2)
(2 votes)
• We aren't simply choosing to use one instead of the other. Sal works through a chain of reasoning to show that in this calculation, all the terms cancel out except for the (n + 1) part of the first term and the (n + 2) term of the last one. Then we see that when we take the limit as n goes to infinity, the first one is unaffected but the second one disappears, so we're simply left with the first (n + 1) term.
(2 votes)
• Sal mentioned at the very end of the video that the series was a telescoping series. Now my question with these types of series is that if you had a telescoping series like the one in the video where it’s the summation from n = 1, to n = infinity, wouldn’t the answer always just be the first term of the partial fraction decomposition?

What I mean is that you would only need to find the first fraction of the partial fraction. Like A/(n+1) was the answer for this video.
(1 vote)
• In the special case of these telescoping series, all you are left with when you take the infinite limit is the first term. But beware! Do not fall into the trap of thinking that any infinite series evaluates to the first term of its partial fraction decomposition. That is most likely not true for a given series.
(4 votes)
• how did u cancel out -2/(N+1)
(2 votes)

## Video transcript

So what we're going to attempt to do is evaluate this sum right over here, evaluate what this series is, negative 2 over n plus 1 times n plus 2, starting at n equals 2, all the way to infinity. And if we wanted to see what this looks like, it starts at n equals 2. So when n equals 2, this is negative 2 over 2 plus 1, which is 3, times 2 plus 2, which is 4. Then when n is equal to 3, this is negative 2 over 3 plus 1, which is 4, times 3 plus 2, which is 5. And it just keeps going like that, negative 2 over 5 times 6. And it just keeps going on and on and on. And now, it looks pretty clear that each successive term is getting smaller. And it's getting smaller reasonably fast. So it's reasonable to assume that even though you have an infinite number of terms, it actually might give you a finite value. But it doesn't jump out at me, at least the way that I've looked at it right now, as to what this sum would be, or how to actually figure out that sum. So what I want you to do now is pause this video. And I'm going to give you a hint about how to think about this. Try to dig up your memories of partial fraction expansion, or partial fraction decomposition, to turn this expression into the sum of two fractions. And that might help us think about what this sum is. So I'm assuming you've given a go at it. So let's try to manipulate this thing. And let's see if we can rewrite this as a sum of two fractions. So this is negative 2 over-- and I'm going to do this in two different colors-- n plus 1 times n plus 2. And we remember from our partial fraction expansion that we can rewrite this as the sum of two fractions, as A over n plus 1 plus B over n plus 2. And why is this reasonable? Well, if you're adding two fractions, you want to find a common denominator, which would be a multiple of the two denominators. This is clearly a multiple of both of these denominators. And we learned in partial fraction expansion that whatever we have up here, especially because the degree here is lower than the degree down here, whatever we have up here is going to be a degree lower than what we have here. So this is a first-degree term in terms of n, so these are going to be constant terms up here. So let's figure out what A and B are. So if we perform the addition-- well, let's just rewrite both of these with the same common denominator. So let's rewrite A over n plus 1, but let's multiply the numerator and denominator by n plus 2. So we multiply the numerator by n plus 2 and the denominator by n plus 2. I haven't changed the value of this first fraction. Similarly, let's do the same thing with B over n plus 2. Multiply the numerator and the denominator by n plus 1, so n plus 1 over n plus 1. Once again, I haven't change the value of this fraction. But by doing this, I now have a common denominator, and I can add. So this is going to be equal to n plus 1 times n plus 2 is our denominator. And then our numerator-- let me expand it out. This is going to be, if I distribute the A, it is An plus 2A. So let me write that, An plus 2A. And then let's distribute this B, plus Bn plus B. Now, what I want to do is I want to rewrite this so I have all of the n terms. So for example, An plus Bn-- I could factor an n out. And I could rewrite that as A plus B times n, those two terms right over there. And then these two terms, the 2A plus B, I could just write it like this, plus 2A plus B. And, of course, all of that is over n plus 1 times n plus 2. So how do we solve for A and B? Well, the realization is that this thing must be equal to negative 2. These two things must be equal to each other. Remember, we're making the claim that this, which is the same thing as this, is equal to this. That's the whole reason why we started doing this. So we're making the claim that these two things are equivalent. We're making this claim. So everything in the numerator must be equal to negative 2. So how do we do that? It looks like we have two unknowns here. To figure out two unknowns, we normally need two equations. Well, the realization here is, look, we have an n term on the left-hand side here. We have no n term here. So you literally could view this, instead of just viewing this as negative 2, you could view this as negative 2 plus 0n, plus 0 times n. That's not "on." That's 0-- let me write it this way-- 0 times n. So when you look at it this way, it's clear that A plus B is the coefficient on n. That must be equal to 0. A plus B must be equal to 0. And this is kind of bread-and-butter partial fraction expansion. We have other videos on that if you need to review that. And the constant part, 2A plus B, is equal to negative 2. And so now we have two equations in two unknowns. And we could solve it a bunch of different ways. But one interesting way is let's multiply the top equation by negative 1. So then this becomes negative A minus B is equal to-- well, negative 1 times 0 is still 0. Now we can add these two things together. And we are left with 2A minus A is A, plus B minus B-- well, those cancel out. A is equal to negative 2. And if A is equal to negative 2, A plus B is 0, B must be equal to 2. Negative 2 plus 2 is equal to 0. We solved for A. And then I substituted it back up here. So now we can rewrite all of this right over here. We can rewrite it as the sum-- and actually, let me do a little bit instead. Let me just write it as a finite sum as opposed to an infinite sum. And then we can just take the limit as we go to infinity. So let me rewrite it like this. So this is the sum from n equals 2-- instead to infinity, I'll just say to capital N. And then later, we could take the limit as this goes to infinity of-- well, instead of writing this, I can write this right over here. So A is negative 2. So it's negative 2 over n plus 1. And then B is 2, plus B over n plus 2. So once again, I've just expressed this as a finite sum. Later, we can take the limit as capital N approaches infinity to figure out what this thing is. Oh, sorry, and B-- let me not write B anymore. We now know that B is 2 over n plus 2. Now, how does this actually go about helping us? Well, let's do what we did up here. Let's actually write out what this is going to be equal to. This is going to be equal to-- when n is 2, this is negative 2/3, so it's negative 2/3, plus 2/4. So that's n equals-- let me do it down here, because I'm about to run out of real estate. That is when n is equal to 2. Now, what about when n is equal to 3? When n is equal to 3, this is going to be negative 2/4 plus 2/5. What about when n is equal to 4? I think you might see a pattern that's starting to form. Let's do one more. When n is equal to 4, well, then, this is going to be negative 2/5-- let me do that same blue color-- negative 2/5 plus 2/6. And we're just going to keep going. Let me scroll down to get some space-- we're going to keep going all the way until the N-th term. So plus dot dot dot plus our capital N-th term, which is going to be negative 2 over capital N plus 1 plus 2 over capital N plus 2. So I think you might see the pattern here. Notice, from our first when n equals 2, we got the 2/4. But then when n equals 3, you had the negative 2/4. That cancels with that. When n equals 3, you had 2/5. Then that cancels when n equals 4 with the negative 2/5. So the second term cancels with-- the second part, I guess, for each n, for each index, cancels out with the first part for the next index. And so that's just going to keep happening all the way until n is equal to capital N. And so this is going to cancel out with the one right before it. And all we're going to be left with is this term and this term right over here. So let's rewrite that. So we get-- let's get more space here. This thing can be rewritten as the sum from lowercase n equals 2 to capital N of negative 2 over n plus 1 plus 2 over n plus 2 is equal to-- well, everything else in the middle canceled out. We're just left with negative 2/3 plus 2 over capital N plus 2. So this was a huge simplification right over here. And remember, our original sum that we wanted to calculate, that just has a limit as capital N goes to infinity. So let's just take the limit as capital N goes to infinity. So let me write it this way. Well, actually, let me write it this way. The limit-- so we can write it this way. The limit as capital N approaches infinity is going to be equal to the limit as capital N approaches infinity of-- well, we just figured out what this is. This is negative 2/3 plus 2 over capital N plus 2. Well, as n goes to infinity, this negative 2/3 doesn't get impacted at all. This term right over here, 2 over an ever larger number, over an infinitely large number-- well, that's going to go to 0. And we're going to be left with negative 2/3. And we're done. We were able to figure out the sum of this infinite series. So this thing right over here is equal to negative 2/3. And this type of series is called a telescoping series-- telescoping, I should say. This is a telescoping series. And a telescoping series is a general term. So if you were to take its partial sums, it has this pattern right over here, where, in each term, you're starting to cancel things out. So what you're left with is just a fixed number of terms at the end. But either way, this was a pretty-- it's a little bit hairy, but it was a pretty satisfying problem.