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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" 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 ?
• 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.
• 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.
• 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?
• 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.
• <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)