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## Integral Calculus

### Course: Integral Calculus>Unit 5

Lesson 17: Telescoping series

# Divergent 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 take a close look at the series 1-1+1-1+1-... Created by Sal Khan.

## Want to join the conversation?

• I recommend watching this Numberphile video to further understand Grandi's series, the one Sal uses as an example. http://www.youtube.com/watch?v=PCu_BNNI5x4
Correct me if I'm wrong but Sal saying that this sum "doesn't exist" in is incorrect. This is a divergent series but there are methods to assign a value to this series so it must "exist". It's definitely "real". •  Numberphile is incorrect, or at least sloppy.

The method in question is called a Cesàro summation. If, and only if, the infinite series absolutely converges will the Cesàro summation be the same as the actual sum. If the series does not absolutely converge, it may still have a Cesàro sum, but not an actual sum -- which means the Cesàro sum might be a curiosity, but not an meaningful sum.

There are other ways of summing infinite series that can be used to "prove" that infinite series sum to clearly false things. For example, you can use a particular method to "prove" that the sum of all positive integers is − 1/12. Obviously, the actual sum approaches infinity, but this supposed "proof" is bandied about in mathematical circles sometimes.
• How is this calculation different from an Eigenvalue. I believe the Eigenvalue for 1+1-1+1-1.....ect is 1/2. • No, an Eigenvalue is a scalar value associated with matrix math.

The term you are looking for, I think, is a Cesàro summation. This is a way of assigning a value to an infinite sum that has no true sum (in math language, a divergent series). Only a special group of divergent series can have a Cesàro summation assigned to them − amongst other things the series cannot diverge to either +∞ or −∞. But, if it is the case that as you add more and more terms to a divergent series, the partial sums AVERAGE out to a particular finite number, then that average is assigned as its Cesàro summation.

So, the series 1 − 1 + 1 − 1.... has partial sums that alternate between 1 and 0, so this series diverges and has no sum. However, since it bounces between two finite numbers, we can just average those numbers and say that, on average, it is ½.

It is important to understand that Cesàro summation is an ASSIGNED value, it is NOT a true sum. As such it can be useful for computations where an AVERAGE value is acceptable, but not in computations where you need the true sum.

For reference sake, the failure to distinguish between a true value and an assigned value (including this very Cesàro summation) was used in a Numberphile video on Youtube to allegedly "prove" that the sum of all positive integers is −1/12. This alleged "proof" also used the associative property where it was not valid. (This alleged "proof" has been around for at least a century, so Numberphile did not invent it.)
• Why does this series diverge ? Just because it doesn't converge ? I thought diverging means to be separating or growing apart or heading to infinity ? So is an oscillating series like the present one which is confined to oscillate between 2 bounds really said to be diverging ? • There are several answers to this question. 1-(1+1)-(1+1)... would approach negative infinity: 1-2-2-2-2-2...
-1 could be written as +(-1) so 1+(-1+1)+(-1+1) would be 1+0+0... = 1
(1-1)+(1-1)+(1-1)...=0+0+0=0
You could also group it in other different ways. So I don't understand this video. The series could diverge or converge. • so this means that either a series converges, or it diverges, there can be no middle no convergence/no divergence situation? • <GIVEN>

Let:

` ∞`
`S = ∑ a(n)`
` n=1`

be an infinite series such that:

`S(n) = 8n/(n + 5)`

Find:

`11`
` ∑ a(n)`
`n=1`

</GIVEN>

<HINTS>

`11`
` ∑ a(n)` is the same as `S(11)`.
`n=1`

`S(11) = 8*11/(11 + 5) = 8*11/16 = 11/2`

</HINTS>

How is a finite summation of a series equal to single index evaluation?

`n 8n n+5 8n/(n + 5)`
`1 8 6 4/3`
`2 16 7 16/7`
`3 24 8 3/1`
`4 32 9 32/9`
`5 40 10 4/1`
`6 48 11 48/11`
`7 56 12 14/3`
`8 64 13 64/13`
`9 72 14 36/7`
`10 80 15 16/3`
`11 88 16 11/2`

Find new common denominator (CD) by computing the LCM for:

`2,3,7,11,13:`

`n 2 3 3 7 11 13`
`2 1 1 1 1 1 1`
`3 3 1 1 1 1 1`
`7 7 7 7 1 1 1`
`9 9 3 1 1 1 1`
`11 11 11 11 11 1 1`
`13 13 13 13 13 13 1`

`CD = LCM = 2*3*3*7*11*13`

`CD = LCM = 18018`

`8n/(n + 5) LCM/D N*LCM/D 4/3 18018/3 = 6006 4*6006 = 2402416/7 18018/7 = 2574 16*2574 = 41184 3/1 18018/1 = 18018 3*18018 = 5405432/9 18018/9 = 2002 32*2002 = 64064 4/1 18018/1 = 18018 4*18018 = 7207248/11 18018/11 = 1638 48*1638 = 7862414/3 18018/3 = 6006 14*6006 = 8408464/13 18018/13 = 1386 64*1386 = 8870436/7 18018/7 = 2574 36*2574 = 9266416/3 18018/3 = 6006 16*6006 = 9609611/2 18018/2 = 9009 11*9009 = 99099 ───── 794669`

`11`
` ∑ = 794669/18018 ≈ 44.1042`
`n=1` • Remember that there are many functions that can be written both as a epsilon sum and as a single evaluation function. (If you review the section about Taylor series you can see how).
In this instance S(n) is the formula for the partial sum of the given epsilon sum, so S(n)=a(1)+a(2)+a(3)+...+a(n), where the "a" are the individual terms of the sum.
• I think that what Sal said (and forgive me for this) at is incorrect. Using algebraic manipulation, you can achieve the answer of 1/2. I am very torn as I have seen a Cesàro proof for this and Sal just said it was diverging. The answer 1/2 is correct, I know, but why would he say that it wasn't? • There are different types of summation of series. According to traditional summation, the series diverges since the sequence of partial sums 1,0,1,0,1,0,... does not converge to a limit. However, according to Cesaro summation (which is defined as taking the limit as n goes to infinity of the average of the first n partial sums), the sum of the series would be 1/2 (since the sequence 1, 1/2, 2/3, 2/4, 3/5, 3/6, 4/7, 4/8,... converges to 1/2).
In other words, the series 1-1+1-1+... diverges, but if we decide to go "out of the box" and assign this series a sum anyway, the most reasonable sum would be 1/2.
Have a blessed, wonderful day!
(1 vote)
• I must be missing something in this progression of videos -

What is the distinction between s_n and the sum of a_n?

Why are we allowed to take the limit as n->infinity of the former to solve the sum, but not the latter?

When using partial fraction decomposition, why can we not use properties of sums to evaluate the fractions separately? For example, the sum (-2/n-1 + 2/n+1) is equal to the sum (-2/n-1) plus the sum(2/n+1) • Hello, everyone. This is a bit of an off topic question, but I want to be a physicist, and to exceed in my field I have to be knowledgable in many areas of mathematics and calculus. So my question is, what resources and order would you recommend I go on so I can succeed in my field? Thanks!
(1 vote) • If you want details on order of physics courses, I would talk to your academic advisor.
But if you want an interesting approach to calculus, try Calculus for the Practical Man by J.E. Thompson. Richard Feynman used this book, and this allowed him to solve problems that stumped others, since he was naturally approaching it from a different angle than his peers.
(1 vote)
• Though this series doesn't converge, the summation of its infinite terms will be 0.5 right?
(1 vote) ## Video transcript

Lets say that we have the sum one minus one plus one minus one plus one and just keeps going on and on and on like that forever and we can write that with sigma notation. This would be the sum from n is equal to one lower case n equals one to infinity. We have an infinite number of terms here but see this first one, we want it to be a positive one, and then we want to keep switching terms. So we could say that this is negative 1 to the lowercase n minus 1 power. Let's just verify that that works. When n is equal to 1, it's negative 1 to the 0 power, which is that. When n is equal to 2, it's 2 minus 1; it's negative one to the first power that's equal to that right over there. So this is a way of writing this series. Now what I wanna think about is does this series converge to an actual finite value? Or, and this is another way of saying it, what is the sum? Is there a finite sum that is equal to this right over here, or does this series diverge? And the way that we can think about that is by thinking about its partial sums let me write that down. The partial, the partial sums of this series. And the way we can define the partial sums, so we'll give an index here. So capital N, so the partial sum is going to be the sum from n equals one but not infinity but to capital N of negative 1 to the n minus 1. So just to be clear, what this means, so the, the partial sum with just one term is just gonna be from lowercase n equals 1 to uppercase N equals 1. So it's just going to be this first term right over here. It's just going to be 1. The S sub two, s sub two is going to be equal to one minus one. It's gonna be the sum of the first two terms. S sub three, S sub three is going to be one minus one plus one. It's the sum of the first three terms, which is of course equal to. Equal to, let's see this equal to one. This one over here is equal to zero. S sub 4, we could keep going, S sub four is going to be one minus one plus one minus one which is equal to zero again. So once again, the question is, does this sum converge to some finite value? And I encourage you to pause this video and think about it, given what we see about the partial sums right over here. So in order for a series to converge, that means that the limit, an infinite series to converge, that means that the limit, the limit, so if you're a convergence, convergence is the same thing, is the same thing as saying that the limit as capital, the limit as capital N approaches infinity of our partial sums is equal to some finite. Let me just write like this, is equal to some Finite, so Finite Value. So, what is this limit going to be? Well, let's see if we can write this. So, this is going to be, let's see s sub n, if we want to write this in general terms. We already see if s, if capital N is odd, it's equal to 1. If capital N is even, it's equal to 0. So, we can write, lets write this down so s sub n I could write it like this is going to be one if n odd it's equal to zero if n even. So what's the limit as s sub n approaches infinity so what's the limit. What's the limit, as N approaches infinity of S sub N. Well, this limit doesn't exist. It keeps oscillating between these points. You give me, you, you go one more, it goes from 1 to 0. You give me one more, it goes from 0 to 1. So it actually is not approaching a finite value. So this right over here does not exist. It's tempting, because it's bounded. It's only, it keeps oscillating between 1 and 0. But it does not go to one particular value as n approaches infinity. So here we would say that our series s diverges. Our series S diverges.