Sum of series 3.

\(\;\;\displaystyle \sum_{n=1}^{\infty} \frac{1}{(2 n-1)^{2}}\)

SOLUTION:

Add and subtract \(\displaystyle\;\; \sum_{n=1}^{\infty} \frac{1}{(2 n)^{2}}\;\;\) we will get

\(\displaystyle \left(\sum_{n=1}^{\infty} \frac{1}{(2 n-1)^{2}}+\sum_{n=1}^{\infty} \frac{1}{(2 n)^{2}}\right)-\sum_{n=1}^{\infty} \frac{1}{2 n^{2}}\)

\(\Rightarrow \displaystyle \left(\sum_{n=1}^{\infty}\left[\frac{1}{(2 n-1)^{2}}+\frac{1}{(2 n)^{2}}\right]\right)-\sum_{n=1}^{\infty} \frac{1}{(2 n)^{2}}\;\;\;\; \cdots (1)\)

We will use

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2 n-1)^{2}}+\frac{1}{(2 n)^{2}} = \sum_{n=1}^{\infty} \frac{1}{n^{2}}\)

in (1), then we will get

\(\displaystyle \left(1+\frac{1}{2^{2}}+\frac{1}{\left(3)^{2}\right.}+\frac{1}{(4)^{2}}+\cdots\right)-\sum_{n=1}^{\infty} \frac{1}{(2 n)^{2}}\)

\(\Rightarrow \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}-\sum_{n=1}^{\infty} \frac{1}{4 n^{2}}\;\;\;\;\cdots (2)\)

We have a result that \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^2}{6}\)

we will use this in (2)

\(\displaystyle \frac{\pi^{2}}{6}-\frac{1}{4}\times \frac{\pi^2}{6} = \frac{\pi^{2}}{6}- \frac{\pi^2}{24}\)

After solving this we will get

\(\displaystyle \frac{\pi^2}{8}\)

ANSWER : \(\displaystyle \frac{\pi^2}{8}\)