Sum of alternate terms of Riemann Zeta function

Question

If $\sum\limits_{n=1}^{\infty}\frac{1}{n^{4}}=\frac{\pi^{4}}{90}$
Then find the value of $\sum\limits_{n=1}^{\infty}\frac{1}{(2n-1)^{4}}$
The book I took this problem from makes no mention of Riemann Zeta function at all. In fact, the book had problems only on arithmetic, geometric and harmonic progressions. This is neither one of them. Any hints?

Answer 1

In general we have for all $n\ge 1$ $$ \zeta(n)=\frac{2^n}{2^n-1}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^n} $$ This can be proved as follows \begin{align*} \zeta(n) & = \left(\frac{1}{1^n}+\frac{1}{3^n}+\cdots\right)+\left(\frac{1}{2^n}+\frac{1}{4^n}+\cdots\right) \cr & =\left(\frac{1}{1^n}+\frac{1}{3^n}+\cdots\right)+\frac{1}{2^n}\zeta(n) \cr & = \left(1+\frac{1}{2^n}+\frac{1}{2^{2n}}+\cdots\right)\sum_{k=1}^{\infty} \frac{1}{(2k-1)^n}. \end{align*} Now use the geometric series with $q=\frac{1}{2^n}$.

Answer 2

Hint: Break up the sum into even and odd parts. $$\sum_{i=1}^\infty\frac{1}{n^4} = \sum_{i=1}^\infty\frac{1}{(2n-1)^4}+\sum_{i=1}^\infty\frac{1}{(2n)^4}$$ You know the value of the quantity on the LHS, and can easily find the value of $\sum_{i=1}^\infty\frac{1}{(2n)^4}$. Solve for the remaining quantity and you have your answer.