Sum of complex series

• June 26th 2007, 05:02 AM
moolimanj
Sum of complex series
Determine the sum of the series:

$\sum_{n=1}^{\infty}(-1)^n/(36n^2-1)$
• July 13th 2007, 12:55 PM
ThePerfectHacker
Quote:

Originally Posted by moolimanj
Determine the sum of the series:

$\sum_{n=1}^{\infty}(-1)^n/(36n^2-1)$

A little late.

Define the function,
$g(z) = \pi f(z) \csc \pi z = \frac{\pi \csc \pi z}{36z^2 - 1}$

Then,
$\sum_{n=-\infty}^{\infty}(-1)^n f(n) = -\sum_{k=1}^{m} \mbox{res}(g,a_k)$

Where $a_1,a_2,...,a_m$ are the poles of $f(z) = \frac{1}{36z^2 - 1} = \frac{1}{36\left(z - \frac{1}{6}\right)\left( z + \frac{1}{6} \right)}$

Thus, $a_{1,2} = \pm \frac{1}{6}$

Then,
$\mbox{res}(g,\pm 1/6) = \lim_{z\to \pm 1/6} \left(z \mp \frac{1}{6} \right) \cdot \frac{\pi \csc \pi z}{36 \left( z - \frac{1}{6} \right) \left( z + \frac{1}{6} \right)}$

Thus,
$\sum_{k=1}^2 \mbox{res}(g,\pm 1/6) = - \frac{\pi}{12}$.

Which means,
$\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{36n^2 - 1} = \frac{\pi}{12}$

Thus,
$2\sum_{n=0}^{\infty} \frac{(-1)^n}{36n^2 - 1} = \frac{\pi}{12}$

Thus,
$\sum_{n=0}^{\infty} \frac{(-1)^n}{36n^2-1} = \frac{\pi}{24}$