Answer
See proof
Work Step by Step
Series: 1+$\frac{1}{3^{4}}$ + $\frac{1}{5^{4}}$+$\frac{1}{7^{4}}$+ ...... is equivalent to $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{4}}$
which is equal to $\sum_{n=1}^{\infty} \frac{1}{(n)^{4}} - \sum_{n=1}^{\infty} \frac{1}{(2n)^{4}}$
=$\sum_{n=1}^{
\infty} \frac{1}{(n)^{4}}-\sum_{n=1}^{\infty} \frac{1}{2^{4}\cdot n^{4}}$
= $\frac{\pi^{4}}{90}$ - $\frac{\pi^{4}}{2^{4}\cdot 90}$
=$\frac{\pi^{4}}{96}$
