Show that $\sum_{n=1}^\infty \frac{n^2}{(n+1)!}=e-1$

Question

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.

Answer 1

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

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

Answer 2

$$\frac{n^2}{(n+1)!}=\frac{(n+1)^2-2(n+1)+1}{(n+1)!}=\frac{n+1}{n!}-2\frac{1}{n!}+\frac{1}{(n+1)!}=\frac{1}{(n-1)!}-\frac{1}{n!}+\frac{1}{(n+1)!}$$ so $$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=\sum^\infty_{n=1}\frac{1}{(n-1)!}-\sum^\infty_{n=1}\frac{1}{n!}+\sum^\infty_{n=1}\frac{1}{(n+1)!}=(e)-(e-1)+(e-2)=e-1$$

Answer 3

$$\frac{n^2}{(n+1)!} = \frac{(n+1)(n-1) + 1}{(n+1)!} = \frac{(n-1)}{n!} + \frac{1}{(n+1)!}$$

Remembering that we're summing to infinity, evaluating the first terms and paying careful attention to the indices,

$$ \begin{align} \sum_{n=1}^\infty \left( \frac{(n-1)}{n!} + \frac{1}{(n+1)!} \right) &= \sum_{n=2}^\infty \frac{(n-1)}{n!} + \sum_{n=2}^\infty \frac{1}{n!}\\ &= \sum_{n=2}^\infty \frac{n}{n!}\\ &=\left( \sum_{n=1}^\infty \frac{n}{n!} \right) - 1\\ &= e - 1 \end{align} $$