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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.

@christianp This looks like a good candidate for induction/limit. I would move to partial sums, then pull the sum inside the product for what is here the denominator, and work out a formula in terms of the number of terms n in the partial sum, then take the limit of that formula as n increases.

@christianp *frowns* ok, the sum doesn't pull inside /directly/, but you get what i mean; it's only a five term expansion at worst. Definitely go from (inductively proved) finite sum formulas on the individual sub-functions of k.

@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).

@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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