Jun 10, 2021 - 7:34

An infinite series problem.

By Vamshi Jandhyala in mathematics

June 10, 2021

Problem

Solution

From Euler’s solution of the Basel Problem, we have

$$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $$

Also,

$$ \begin{aligned} \sum_{n=1}^\infty \frac{1}{n^2} &= \sum_{n=1}^\infty \frac{1}{(2n)^2} + \frac{1}{(2n+1)^2} \\
\implies \frac{\pi^2}{6} &= \frac{1}{4} \cdot \frac{\pi^2}{6} + \sum_{n=1}^\infty \frac{1}{(2n+1)^2}\\
\implies \sum_{n=1}^\infty \frac{1}{(2n+1)^2} &= \frac{\pi^2}{8} \end{aligned} $$

Therefore,

$$ \begin{aligned} &1 + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{11^2} + \frac{1}{13^2} + \dots \\
&= \sum_{n=1}^\infty \frac{1}{n^2} - \frac{1}{2^2}\sum_{n=1}^\infty \frac{1}{n^2} - \frac{1}{3^2}\sum_{n=1}^\infty \frac{1}{(2n+1)^2} \\
&= \frac{\pi^2}{6} - \frac{\pi^2}{24} - \frac{\pi^2}{72} \\
&= \frac{\pi^2}{9} \end{aligned} $$