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.
 A: You have almost done it,note that $$\sum _{ n=1 }^{ \infty  } \frac { n^{ 2 } }{ (n+1)! } =\sum _{ n=1 }^{ \infty  } \frac { n^{ 2 }-1+1 }{ (n+1)! } =\sum _{ n=1 }^{ \infty  } \left( \frac { n-1 }{ n! } +\frac { 1 }{ (n+1)! }  \right) =\sum _{ n=1 }^{ \infty  } \left( \frac { 1 }{ \left( n-1 \right) ! } -\frac { 1 }{ n! } +\frac { 1 }{ (n+1)! }  \right) =$$
here $$\sum _{ n=1 }^{ \infty  } \left( \frac { 1 }{ \left( n-1 \right) ! } -\frac { 1 }{ n! }  \right) =1$$ is telescoping series so

$$\sum _{ n=1 }^{ \infty  } \left( \frac { 1 }{ \left( n-1 \right) ! } -\frac { 1 }{ n! } +\frac { 1 }{ (n+1)! }  \right) =1+\left( e-2 \right) = \color {blue}{e-1}$$

A: $$\frac{n^2}{(n+1)!} = \frac{(n+1)(n-1) + 1}{(n+1)!} = \frac{(n-1)}{n!} + \frac{1}{(n+1)!}$$
Remembering that we're summing to infinity, evaluating the first terms and  paying careful attention to the indices,
$$
\begin{align}
\sum_{n=1}^\infty \left( \frac{(n-1)}{n!} + \frac{1}{(n+1)!} \right) &= \sum_{n=2}^\infty \frac{(n-1)}{n!} + \sum_{n=2}^\infty \frac{1}{n!}\\ &= \sum_{n=2}^\infty \frac{n}{n!}\\ &=\left( \sum_{n=1}^\infty \frac{n}{n!} \right) - 1\\
&= e - 1
\end{align} $$
A: $$\frac{n^2}{(n+1)!}=\frac{(n+1)^2-2(n+1)+1}{(n+1)!}=\frac{n+1}{n!}-2\frac{1}{n!}+\frac{1}{(n+1)!}=\frac{1}{(n-1)!}-\frac{1}{n!}+\frac{1}{(n+1)!}$$
so
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=\sum^\infty_{n=1}\frac{1}{(n-1)!}-\sum^\infty_{n=1}\frac{1}{n!}+\sum^\infty_{n=1}\frac{1}{(n+1)!}=(e)-(e-1)+(e-2)=e-1$$
A: 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.
