Telescoping Series theorem vs. Grandi's series

  • Thread starter marenubium
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  • #1

Homework Statement


No actual problem, thinking about the telescoping series theorem and Grandi's series

For reference Grandi's series S = 1 - 1 + 1 - 1...

Homework Equations


[/B]
The telescoping series theorem in my book states that a telescoping series of the form (b1 - b2) + ... + (bn - bn+1) + ... converges IFF limn->inf bn exists.

S can be written (1 - 1) + (1 - 1) + ...
so following the template of the telescoping series theorem, bn = 1.

The Attempt at a Solution



Since limn->inf bn = 1 then by the telescoping series theorem S converges.

If I think about S as a sequence of partial sums, specifically
S1 = 1
S2 = 1 - 1 = 0
S3 = 1 - 1 + 1 = 1
...

then this sequence diverges and thus S diverges. I don't know how to resolve this contradiction.

I'm really really sorry if this is a common question. I googled this, did a search of this forum, and read about Grandi's series on wikipedia. I haven't been able to find an explanation that I can absorb that satisfies me.
 

Answers and Replies

  • #2
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The wikipedia article talks about the history of the Grandi series and how it fueled debates for a long time. There's also a discussion in the article on divergence and partial sums that shows that the series is in fact divergent.

In modern mathematics, the sum of an infinite series is defined to be the limit of the sequence of its partial sums, if it exists. The sequence of partial sums of Grandi's series is 1, 0, 1, 0, ..., which clearly does not approach any number (although it does have two accumulation points at 0 and 1). Therefore, Grandi's series is divergent.

It can be shown that it is not valid to perform many seemingly innocuous operations on a series, such as reordering individual terms, unless the series is absolutely convergent. Otherwise these operations can alter the result of summation.[5] Further, the terms of Grandi's series can be rearranged to have its accumulation points at any interval of two or more consecutive integer numbers, not only 0 or 1. For instance, the series

## 1+1+1+1+1-1-1+1+1-1-1+1+1-1-1+1+1##

(in which, after five initial +1 terms, the terms alternate in pairs of +1 and −1 terms) is a permutation of Grandi's series in which each value in the rearranged series corresponds to a value that is at most four positions away from it in the original series; its accumulation points are 3, 4, and 5.

from the article: https://en.wikipedia.org/wiki/Grandi's_series

Here's a tutorial with some discussion on telescoping series and convergence that may help:

http://tutorial.math.lamar.edu/Classes/CalcII/Series_Special.aspx

They mention that the limit of partial sums must exist for the series to converge not the limit of the terms of the telescoping series. In your case, the partial sums alternates between 0 and 1 and hence doesn't exist.

Perhaps @Mark44 or @fresh_42 can provide a better answer to this most interesting question.
 
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  • #3
I guess, to succinctly word my question, doesn't this result (the Grandi's series diverges) contradict the conclusion of the Telescoping Series Theorem, as applied to this series?
 
  • #4
12,935
6,778
I guess, to succinctly word my question, doesn't this result (the Grandi's series diverges) contradict the conclusion of the Telescoping Series Theorem, as applied to this series?

I think the key is the limit of the sequence of ##b_{n}## as n approaches infinity doesn't exist it oscillates between -1 and 1 depending on whether ##n## is even or odd.
 
  • #5
Ray Vickson
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Homework Statement


No actual problem, thinking about the telescoping series theorem and Grandi's series

For reference Grandi's series S = 1 - 1 + 1 - 1...

Homework Equations


[/B]
The telescoping series theorem in my book states that a telescoping series of the form (b1 - b2) + ... + (bn - bn+1) + ... converges IFF limn->inf bn exists.

If your book really says that the series converges if and only if ##\lim_{n \to \infty} b_n## exists, then your book is wrong. In fact, the series converges if and only if ##\lim_{n \to \infty} b_n = 0##.

That is easy to see: for the partial sum ending with ##(b_{n-1} - b_n)## the value of the sum is ##b_1 - b_n##. For the partial sum with one more term the sum is equal to ##b_1 + \cdots +(b_{n-1} - b_n) + b_n = b_1##. Thus, the partial sums alternate between ##b_1## and ##b_1 -b_n##, so have a limit as ##n \to \infty## if, and only if ##b_n \to 0##.
 
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