TITLE: Evaluate the series: $ \sum_{k=1}^{\infty}\frac{1}{k(k+1)^2k!}$
QUESTION [7 upvotes]: Evaluate the series:
$$ \sum_{k=1}^{\infty}\frac{1}{k(k+1)^2k!}$$

REPLY [5 votes]: Try it with factor $x^k$.
$$\begin{align}
f(x) &=\sum_{k=1}^{\infty}\frac{x^k}{k(k+1)^2k!}
\\
f'(x) &=\sum_{k=1}^{\infty}\frac{x^{k-1}}{(k+1)^2k!}
\\
x^2f'(x) &=\sum_{k=1}^{\infty}\frac{x^{k+1}}{(k+1)^2k!}
\\
\big(x^2f'(x)\big)' &=\sum_{k=1}^{\infty}\frac{x^{k}}{(k+1)k!}
\\
x\big(x^2f'(x)\big)' &=\sum_{k=1}^{\infty}\frac{x^{k+1}}{(k+1)k!}
\\
\Big(x\big(x^2f'(x)\big)'\Big)' &=\sum_{k=1}^{\infty}\frac{x^{k}}{k!} = e^x-1 .
\end{align}$$
Now solve a differential equation.  Then plug in $x=1$.