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.

olligobber@olligobber@mathstodon.xyz@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).