Evaluate the series: $ \sum_{k=1}^{\infty}\frac{1}{k(k+1)^2k!}$ Evaluate the series:
$$ \sum_{k=1}^{\infty}\frac{1}{k(k+1)^2k!}$$
 A: 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 $x=1$.
A: Partial fraction decomposition gives
$$\frac{1}{k(k+1)^2}=\left(\frac{1}{k}-\frac{1}{k+1}\right)\frac{1}{k+1}=\frac{1}{k}-\frac{1}{k+1}-\frac{1}{(k+1)^2}$$
Hence this series is
$$\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+1}-\frac{1}{(k+1)^2}\right)\frac{1}{k!}$$
$$=\left(\sum_{k=1}^\infty\frac{1}{k \cdot k!}\right)-\left(\sum_{k=1}^\infty\frac{1}{(k+1)!}\right)-\left(\sum_{r=2}^\infty\frac{1}{r\cdot r!}\right).$$
Notice how in the third sum we set $r=k+1$ so $(k+1)^2k!=(k+1)\cdot(k+1)!=r\cdot r!$. The middle term is clearly $e-2$, and the difference between the outside series is $\frac{1}{1\cdot 1!}$, hence we obtain $3-e$.
