<s>[INST] Evaluate the series: $ \sum_{k=1}^{\infty}\frac{1}{k(k+1)^2k!}$ Evaluate the series:
$$ \sum_{k=1}^{\infty}\frac{1}{k(k+1)^2k!}$$
 [/INST] 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 $3-e$.
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