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### 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" in4:25is 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".(15 votes)- 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.(36 votes)

- How is this calculation different from an Eigenvalue. I believe the Eigenvalue for 1+1-1+1-1.....ect is 1/2.(4 votes)
- 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.)(8 votes)

- 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 ?(3 votes)
- A infinite series converges if and only if it approaches a single finite value. If the infinite series does
**anything**else, it is divergent.

So, by definition, an infinite series is divergent if it is not convergent.(7 votes)

- 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.(3 votes)- The first line is a mistake with parentheses. 1-1+1-1+1... is not equal to 1-(1+1)-(1+1)... because you've changed the sign of the 3rd term (and the 5th, etc.) Your second paragraph corrects this mistake.

But I agree with you, it is tempting to group the series as 1+(-1+1)... or (1-1)+..., and I wish the video had explained why those are not valid ways to determine convergence and the sum of the series.(1 vote)

- so this means that either a series converges, or it diverges, there can be no middle no convergence/no divergence situation?(2 votes)
- That is correct. A series could diverge for a variety of reasons: divergence to infinity, divergence due to oscillation, divergence into chaos, etc. The only way that a series can converge is if the sequence of partial sums has a unique finite limit. So yes, there is an absolute dichotomy between convergent and divergent series.(4 votes)

- <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 = 24024

16/7 18018/7 = 2574 16*2574 = 41184

3/1 18018/1 = 18018 3*18018 = 54054

32/9 18018/9 = 2002 32*2002 = 64064

4/1 18018/1 = 18018 4*18018 = 72072

48/11 18018/11 = 1638 48*1638 = 78624

14/3 18018/3 = 6006 14*6006 = 84084

64/13 18018/13 = 1386 64*1386 = 88704

36/7 18018/7 = 2574 36*2574 = 92664

16/3 18018/3 = 6006 16*6006 = 96096

11/2 18018/2 = 9009 11*9009 = 99099

─────

794669`11`

`∑ = 794669/18018 ≈ 44.1042`

`n=1`

(2 votes)- 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.(3 votes)

- I think that what Sal said (and forgive me for this) at4:25is 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?(2 votes)
- 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)(2 votes) - 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.