<s>[INST] Find the sum of the infinite series $\sum n(n+1)/n!$ How do find the sum of the series till infinity?
$$ \frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\frac{2+4+6+8}{4!}+\cdots$$
I know that it gets reduced to $$\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}$$
But I don't know how to proceed further.
 [/INST] As you observed $$\sum_{n=1}^{\infty} \frac{n(n+1)}{n!}$$ further reduces to,
$$\begin{align}
&\sum_{n=1}^{\infty} \frac{n(n+1)}{n(n-1)!} \\
=&\sum_{n=1}^{\infty} \frac{n+1}{(n-1)!}\\
=&\sum_{n=1}^{\infty} \frac{n}{(n-1)!}+\frac{1}{(n-1)!}\\
=&\sum_{n=1}^{\infty} \frac{(n-1)}{(n-1)!} + \frac{2}{(n-1)!}\\
=&\sum_{n=1}^{\infty} \frac{1}{(n-2)!} +\sum_{n=1}^{\infty} \frac{2}{(n-1)!},
\end{align}$$  
Both summations equal $e$(euler's number) whose expansion is given by,
$$e=\sum_{n=1}^{\infty} \frac{1}{(n-1)!}.
$$ 
So $$\sum_{n=1}^{\infty} \frac{n(n+1)}{n!} = 3e$$
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