- #1

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## 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?

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- Thread starter center o bass
- Start date

- #1

- 560

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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?

- #2

Gib Z

Homework Helper

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- #3

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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

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----

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

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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?

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