I need to find the sum of $\displaystyle\sum_{n=1}^\infty \frac{1}{n^4}.$
In the solution document it written:
$$\displaystyle\sum_{n=1}^\infty \frac1{n^4} = \sum_{n=1}^\infty \frac1{(2n-1)^4} + \sum_{n=1}^\infty \frac1{(2n)^4}.$$
My question is, how did he get that $\displaystyle\sum_{n=1}^\infty \frac1{n^4} = \sum_{n=1}^\infty \frac1{(2n-1)^4} + \sum_{n=1}^\infty \frac1{(2n)^4} $?
Thanks in advance.
\begin{align*} \sum_{n=1}^\infty \frac{1}{n^4} &= \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \frac{1}{5^4} + \frac{1}{6^4} + \cdots \\ &= \left( \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots \right) + \left( \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \cdots \right) \\ &= \sum_{n=1}^\infty \frac{1}{(2n-1)^4} + \sum_{n=1}^\infty \frac{1}{(2n)^4} \end{align*}
Note that $n$ is just a dummy variable, so it need not be true that $\displaystyle \frac{1}{n^4} = \frac{1}{(2n-1)^4} + \frac{1}{(2n)^4}$.
Since the even and odd natural numbers form a complete residue system modulo two the author can separate the sum into two sums of even and odd terms. This could be done with any modulus for example instead of modulo two, we could separate the sum using a modulus of three:
$$\sum_{n=1}^\infty\frac{1}{n^4}=\sum_{n=1}^\infty\frac{1}{(3n)^4}+\sum_{n=1}^\infty\frac{1}{(3n-1)^4}+\sum_{n=1}^\infty\frac{1}{(3n-2)^4}$$
Or for any natural number $m$
$$\sum_{n=1}^\infty\frac{1}{n^4}=\sum_{k=0}^{m-1}\sum_{n=1}^\infty\frac{1}{(mn-k)^4}$$
