<s>[INST] Wolfram Alpha can't find an explicit form (or even an explicit form of an approximation of) this innocuous series? I suppose my question could be restated as "why?" to the above statement. 
I'll rephrase my question if this breaks any rules asking it this way, but the expression is simply:
$$\sum_{m=2}^\infty \frac{1}{m^{2n}(m^2-1)} = ??$$
with the partial sum being:
$$\sum_{m=2}^a \frac{1}{m^{2n}(m^2-1)} = ??$$
$m, n \in \mathbb N$
Are the above two somehow more insidious than they look?
 [/INST] $$
\begin{align}
\sum_{m=2}^\infty\frac1{m^{2n}(m^2-1)}
&=\sum_{m=2}^\infty\sum_{k=n+1}^\infty m^{2k}\\
&=\sum_{k=n+1}^\infty\sum_{m=2}^\infty m^{2k}\\
&=\sum_{k=n+1}^\infty(\zeta(2k)-1)\\
&=\bbox[5px,border:2px solid #C0A000]{\frac34-\sum_{k=1}^n(\zeta(2k)-1)}\tag1
\end{align}
$$
since
$$
\begin{align}
\sum_{k=1}^\infty(\zeta(2k)-1)
&=\sum_{k=1}^\infty\sum_{n=2}^\infty\frac1{n^{2k}}\\
&=\sum_{n=2}^\infty\sum_{k=1}^\infty\frac1{n^{2k}}\\
&=\sum_{n=2}^\infty\frac1{n^2-1}\\
&=\frac12\sum_{n=2}^\infty\left(\frac1{n-1}-\frac1{n+1}\right)\\[3pt]
&=\frac34\tag2
\end{align}
$$
In $(1)$, $\zeta(2k)$ is a rational multiple of $\pi^{2k}$, as shown in this answer.
Here are the first several sums:
$$
\begin{array}{c|l}
n&\frac34-\sum\limits_{k=1}^n(\zeta(2k)-1)\\\hline
1&\frac74-\frac{\pi^2}6\\
2&\frac{11}4-\frac{\pi^2}6-\frac{\pi^4}{90}\\
3&\frac{15}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}\\
4&\frac{19}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}-\frac{\pi^8}{9450}\\
5&\frac{23}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}-\frac{\pi^8}{9450}-\frac{\pi^{10}}{93555}\\
6&\frac{27}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}-\frac{\pi^8}{9450}-\frac{\pi^{10}}{93555}-\frac{691\pi^{12}}{638512875}
\end{array}\tag3
$$
</s>