Evaluate Σn=1^∞(n^3/((n+1)!).

Write $n^3$ in terms of factors of the form $(n+1)n(n-1)$ and lower so the terms cancel against the factorial. Indeed $n^3=(n+1)n(n-1)+n$, and $n=(n+1)-1$. Therefore $\frac{n^3}{(n+1)!}=\frac{(n+1)n(n-1)}{(n+1)!}+\frac{n+1}{(n+1)!}-\frac1{(n+1)!}=\frac{1}{(n-2)!}+\frac1{n!}-\frac1{(n+1)!}$, where the first term is taken as $0$ for $n=1$. Summing, $\sum_{n\ge2}\frac1{(n-2)!}=e$, $\sum_{n\ge1}\frac1{n!}=e-1$ and $\sum_{n\ge1}\frac1{(n+1)!}=e-2$. The total is $e+(e-1)-(e-2)=e+1\approx3.7183$, i.e. $3.72$. Adding the first few terms $0.5+1.333+1.125+0.533+0.174+\dots$ confirms the value. Rounding to two decimal places gives 3.72. The decomposition into falling factorials is the standard way to sum polynomial over factorial series without differentiating the exponential series repeatedly, and the series converges very quickly because of the factorial in the denominator.