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 Unless my early-morning eyes are reading something wrong, the second partial sum is already 1/120 + 4/360 = 7/360 > 1/96.

Well it looks like it does!

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