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 Oh. Part of your formula doesn't display (it's wider than the column, in the three-column display here). So I only see (k+1)*...*(k+4) as the denominator.

@jsiehler @christianp If you open the status in incognito mode (without being logged in) it should be wider. The denominator is (k+1)*...*(k+5).

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