# Sum of Series given another

Given that: $$\sum_{n=1}^\infty \frac{1}{n^4}= \frac{\pi^4}{90}$$ Find the sum of: $$\sum_{n=0}^\infty \frac{1}{(2n+1)^4}$$

$$\sum_{n=0}^\infty \frac{1}{(2n+1)^4}=\sum_{n=1}^\infty \frac{1}{n^4}-\sum_{n=1}^\infty \frac{1}{(2n)^4}$$
It is given that $${\sum_{n=1}^{\infty}}{1\over {n^4}}={{\pi}^2\over 90}$$
We have to find $${\sum_{n=0}^{\infty}}{1\over {(2n+1)^4}}$$ $$={\sum_{n=0}^{\infty}}{1\over {n^4}}-{1\over {(2n)^4}}$$ $$={\sum_{n=0}^{\infty}}{1\over {n^4}}-{1\over 16}{\sum_{n=0}^{\infty}}{1\over {n^4}}$$ $$={{\pi}^2\over 90}\times {15\over 16}$$ $$={{\pi}^2\over 96}$$