How can i solve the series \sum_{n=0}^{\infty} \frac{n^3}{(n+1)!} ?

1 Answer
Jul 27, 2018

Calling f(x) = \frac{e^x-1}{x} = \sum_{k=0}^{\infty}\frac{x^k}{(k+1)!} we have

x\frac{d}{dx}\(x\frac{d}{dx}(x\frac{d}{dx}f(x))\) = \sum_{k=0}^{\infty}k^3 \frac{x^k}{(k+1)!} = \frac{e^x \(x^3+x-1\)+1}{x}

now making x=1 we have

\sum_{k=0}^{\infty} \frac{k^3}{(k+1)!} = e+1