<s>[INST] Show that $\sum_{n=1}^\infty \frac{n^2}{(n+1)!}=e-1$ Show that:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=e-1$$
First I will re-define the sum:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!} = \sum^\infty_{n=1} \frac{n^2-1+1}{(n+1)!} - \sum^\infty_{n=1}\frac{n-1}{n!} + \sum^\infty_{n=1} \frac{1}{(nm)!}$$
Bow I will define e:
$$e^2 = 1+ \frac{2}{1!} + \frac{x^2}{2!} + ... + \infty$$
$$e' = 1 + \frac{1}{1!} + \frac{1}{2!} + ... + \infty$$
$$(e'-2) = \sum^\infty_{n=1} \frac{1}{(n+1)!}$$
Now I need help.
 [/INST] A differential approach follows with by use of $\delta = x D = x \frac{d}{dx}$ and integration. Since $\delta x^n = n x^n$ then
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2 \, x^n}{n!} &= \delta^{2} \, e^{x} = x(x+1) \, e^{x}
\end{align}
now, by integration,
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2 \, t^{n+1}}{(n+1)!} &= \int_{0}^{t} x(x+1) \, e^{x} \, dx = (1-t +t^2) \, e^{t} -1.
\end{align}
This can also be seen in the form
$$\sum_{n=1}^{\infty} \frac{n^2 \, t^{n}}{(n+1)!} = \frac{1 + t^3}{t(1+t)} \, e^{t} - \frac{1}{t}.$$
Setting $t=1$ leads to the desired result.
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