Computing $\sum_{n=1}^{\infty} \frac{1}{n^4}$ I came across this problem in Fourier analysis, but I'm not sure my confusion is with Fourier analysis per se, it might be that I'm lacking some understanding of sums/series.
I am trying to determine the sum $$\sum_{n=1}^{\infty} \frac{1}{n^4}.$$ I know this problem has been posted before, but I want to solve it in a particular way. I already have that $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^4} = \frac{\pi^4}{96}.$$ I found a suggested solution online that said to solve $x = \frac{\pi^4}{96} + \frac{x}{16}$ for $x$ to find the right answer. (So $x$ is the series I am trying to determine.) But I don't understand why I should divide $x$ by $16$. I understand that I have to add something to $\frac{\pi^4}{96},$ since that is the sum $\sum_{n=1}^{\infty} \frac{1}{n^4}$ over all odd indices, but not over the even ones. And I get why I must add $x$ divided by something, but why $16$? 

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By 'comparing' two well known $\ds{\cot}$-expansions:

\begin{align}
\cot\pars{\pi x} & =
{1 \over \pi x} - {1 \over 3}\,\pi x - {1 \over 45}\,\pars{\pi x}^{3} -
{2 \over 945}\,\pars{\pi x}^{5} + \,\mrm{O}\pars{x^{7}}
\\[5mm] & =
\pars{1 \over \pi}{1 \over x} - \pars{\pi \over 3}x -
\pars{\color{#f00}{\pi^{3} \over 45}}x^{3} - \pars{{2\pi^{5} \over 945}}x^{5} + \,\mrm{O}\pars{x^{7}}
\label{1}\tag{1}
\end{align}

\begin{align} 
\cot\pars{\pi x} & =
{1 \over \pi x} + {2 \over \pi}\sum_{n = 1}^{\infty}{x \over x^{2} - n^{2}} =
{1 \over \pi x} -
{2 \over \pi}\sum_{n = 1}^{\infty}{1 \over n^{2}}{x \over 1 - x^{2}/n^{2}}
\\[5mm] & =
{1 \over \pi x} -
\pars{{2 \over \pi}\,\sum_{n = 1}^{\infty}{1 \over n^{2}}}x -
\pars{\color{#f00}{{2 \over \pi}\sum_{n = 1}^{\infty}{1 \over n^{4}}}}x^{3} + \,\mrm{O}\pars{x^{5}}
\label{2}\tag{2}
\end{align}

With \eqref{1} and \eqref{2}:


$\ds{\color{#f00}{\pi^{3} \over 45} =
\color{#f00}{{2 \over \pi}\sum_{n = 1}^{\infty}{1 \over n^{4}}}\quad
\implies\bbx{\ds{\sum_{n = 1}^{\infty}{1 \over n^{4}} = {\pi^{4} \over 90}}}}$

