How does $\frac1{n^4} =\frac1{(2n-1)^4} + \frac1{(2n)^4}$ 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 $\displaystyle \frac{1}{n^4} = \frac{1}{(2n-1)^4} + \frac{1}{(2n)^4}$.
