@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.