Came across the fact that $\frac{1}{96} = \sum_{k=1}^{\infty} \frac{k^2}{(k+1)(k+2)(k+3)(k+4)(k+5)}$, without any references.
How do I go about proving that? I hope the answer doesn't involve the Riemann ζ function.

@christianp It doesn't. First consider the $$n$$-th partial sum. Re-write the numerator as $$k²=k²+9k+20-9k-20=(k+4)(k+5)-9(k+1)-11$$. Use this decomposition to obtain three parts whose limit are easily found by rewritingeach one as a telescope sum. E.g., the first one is $\sum_{k\leq n}\frac{1}{\prod_{i\leq 3}(k+i)}=(1/2)(\sum_{k\leq n}\frac{k+3}{\prod_{i\leq 3}(k+i)}-\sum_{k\leq n}\frac{k+1}{\prod_{i\leq 3}(k+i)}$; next, all except two term cancel; only the former (1/12) survives the limit.

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