<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] 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

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

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