Summation of alternating series

  • #1

Homework Statement


Determine the value of the sum

[tex]\sum_{n=1}^\infty \frac{(-1)^n}{1+n^2}[/tex]

I have determined it trough the use of fourier series, but does there exist another way to do it?
 

Answers and Replies

  • #2
Gib Z
Homework Helper
3,346
6
The summand is an even function in n, so there are standard methods from Complex Analysis that could do it. If you have done some complex analysis and are interested, I could post up some details on how to do it.
 
  • #3
I have done some complex analysis and I did indeed think about that alternative.
I do not have much experience with the complex analysis approach to summation of series though. The only way of summing series trough complex analysis that I've looked at was by considering

[tex]\oint \pi \cot (\pi z) f(z)dz = 2 \pi i \sum_{n=-infty}^{\infty} f(n) [/tex]

where the contour of integration is a square which sides goes to infinity. But the usual requirement is that the integral goes to zero which I couldn't really say that it does when

[tex]f(z) = \frac{\cos (\pi z)}{1+z^2}[/tex].

I would very much like if you could show me how this series is to be summed trough CA :)
 
  • #4
Gib Z
Homework Helper
3,346
6
Ahh silly me, I forgot about about a variation of that method that makes alternating series very easy to deal with. I will post a general statement of results (I'll skip considering the integrals and just give the end result) so you can start working on the sum, and if any of it looks very foreign to you and it's not in your own textbook or notes, I can provide some details.
----

If there exists a [itex] \epsilon > 0 [/itex] such that [tex] |f(z)| = O\left( \frac{1}{|z|^{1+\epsilon}} \right) [/tex] as [itex] |z| \to \infty [/itex], then:

[tex] \sum_{n=-\infty}^{\infty} ' f(n) = - \sum \left( \mbox{ Residues of } \pi f(z) \cot (\pi z) \mbox{ at all the poles of } f(z) \right) [/tex]

[tex] \sum_{n= -\infty}^{\infty} ' (-1)^n f(n) = - \sum \left( \mbox{ Residues of } \pi f(z) \csc (\pi z) \mbox{ at all the poles of } f(z) \right) [/tex]

where the ' after the sums indicates summing over all integers n, except where there is a pole. In this case we are fine, but for example, say we were trying to sum [itex] \sum 1/n^2 [/itex], then with this method we would sum over all integers but omitting 0.

In our problem, it's quite clear that [tex] f(z) = \frac{1}{1+z^2} [/tex] is the function we must consider and it certainly satisfies the growth bound ( with [itex] \epsilon = 1 [/itex]) and so if you apply the formula I provided, it should yield the answer.
 
  • #5
Ah, that got me to the right answer. Thank you! :)

Do you know what kind of contour is considered in the second method with the alternating series?

The first method without alternating series was the one i considered first with
[tex]
f(z) = \frac{\cos (\pi z)}{1+z^2}
[/tex]

since [tex]\frac{(-1)^n}{1+n^2} = \frac{\cos (n \pi)}{1+n^2}[/tex]. But I could'nt really make sence of that since it didnt seem that |f(z)| would go to zero.

Do you think this consideration could also get me there?
 

Related Threads on Summation of alternating series

  • Last Post
Replies
2
Views
514
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
774
  • Last Post
Replies
2
Views
862
  • Last Post
Replies
5
Views
2K
  • Last Post
2
Replies
26
Views
3K
  • Last Post
Replies
7
Views
1K
Top