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\begin{align}
\pi x\cot\pars{\pi x}&=1 + \sum_{n = 1}^{\infty}{2x^{2} \over x^{2} - n^{2}}
=1 - 2x^{2}\sum_{n = 1}^{\infty}{1 \over n^{2}}
- 2x^{4}\sum_{n = 1}^{\infty}{1 \over n^{4}}
- 2x^{6}\sum_{n = 1}^{\infty}{1 \over n^{6}} - \cdots
\end{align}
$$
\pi x^{1/2}\cot\pars{\pi x^{1/2}}=1 - 2x\sum_{n = 1}^{\infty}{1 \over n^{2}}
- 2x^{2}\sum_{n = 1}^{\infty}{1 \over n^{4}}
- 2x^{3}\sum_{n = 1}^{\infty}{1 \over n^{6}} - \cdots
$$
For $\verts{z} \sim 0$:
\begin{align}
z\cot\pars{z}&={z \over \tan\pars{z}} \sim {z \over z + z^{3}/3 + 2z^{5}/15}
={1 \over 1 + z^{2}/3 + 2z^{4}/15}
\\[3mm]&\sim 1 - \pars{{z^{2} \over 3} + {2z^{4} \over 15}}
+ \pars{{z^{2} \over 3} + {2z^{4} \over 15}}^{2}
\sim 1 - {z^{2} \over 3} - {2z^{4} \over 15} + {z^{4} \over 9}
=1 - {z^{2} \over 3} - {z^{4} \over 45}
\end{align}
$$
\pi x^{1/2}\cot\pars{\pi x^{1/2}}\sim
1 - {\pi^{2} \over 3}\,x - {\pi^{4} \over 45}\,x^{2}
$$
$$
\color{#00f}{\large\sum_{n = 1}^{\infty}{1 \over n^{4}}}
= -\,\half\,\pars{-\,{\pi^{4} \over 45}}
= \color{#00f}{\large{\pi^{4} \over 90}}
$$