Problem 1808 (difficulty: 6/10)

\(\displaystyle \frac1{1^3}-\frac1{3^3}+\frac1{5^3}-\frac1{7^3}+\frac1{9^3}-\frac1{11^3}+-\ldots = ? \)

Solution:

Take the sum of residues of \(\displaystyle f(z)=\frac{\frac{\pi}{2}}{\cos \frac{\pi}{2}z} \cdot \frac{1}{z^3}\) in a circle with radius \(\displaystyle 2K\), with a large integer \(\displaystyle K\). Along such circles, \(\displaystyle \frac{\frac{\pi}{2}}{\cos \frac{\pi}{2}z}\) is uniformly bounded, so

\(\displaystyle \left|\sum_{|z|<K\pi} \res_z f \right| < O(1/K^2). \)

Therefore the sum of all residues is \(\displaystyle 0\).

The function has poles at \(\displaystyle 0\) and at all odd integers. The Laurend series around \(\displaystyle 0\) is

\(\displaystyle \frac{\frac{\pi}{2}}{\cos \frac{\pi}{2}z} \cdot \frac{1}{z^3} = \frac{\frac{\pi}{2}}{1 - \frac{\pi^2}{8}z^2 + \ldots} \cdot \frac{1}{z^3} = \frac{\pi/2}{z^3} + \frac{\pi^3/16}{z} + \ldots, \)

so the residue at \(\displaystyle 0\) is \(\displaystyle \frac{\pi^3}{16}\).

At odd integers \(\displaystyle 2k+1\) we have

\(\displaystyle \res_{2k+1} f = \res_{z=2k+1} \left( \frac{\frac{\pi}{2}}{\cos \frac{\pi}{2}z} \cdot \frac{1}{z^3} \right) = \frac{(-1)^k}{(2k+1)^3}. \)

Hence,

\(\displaystyle 0 = \sum \res f = \res_0 + \sum_{k\in\Z} \res_{2k+1} = \frac{\pi^3}{16} + \sum_{k\in\Z} \frac{(-1)^k}{k^3} = \frac{\pi^3}{16} - 2\left(\frac1{1^3}-\frac1{3^3}+\frac1{5^3}-\frac1{7^3}+-\ldots \right), \)

\(\displaystyle \frac1{1^3}-\frac1{3^3}+\frac1{5^3}-\frac1{7^3}+\frac1{9^3}-\frac1{11^3}+-\ldots = \frac{\pi^3}{32}. \)

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