Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$ I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$
$$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}.$$
I know that $$\sum_{n=0}^\infty\frac{1}{n!}=e$$ 
so $$\sum_{n=1}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{n!}-1=e-1$$
But, what can I do for $$\sum_{n=1}^\infty\frac{1}{(n+1)!}$$ ?
Am I allowed to start a sum for $n=-1$ ? How can I bring to a something similar to $$\sum_{n=0}^\infty\frac{1}{n!}$$?
 A: Hint: if $m=n+1$, $\displaystyle \sum_{n=1}^\infty \dfrac{1}{(n+1)!} = \sum_{m=2}^\infty \dfrac{1}{m!}$.
A: Use the telescope rule after your third line, that is:$$\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{n!}} - \frac{1}{{(n + 1)!}}} \right)}  = 1$$ ;)
A: $$\begin{align}
\text{Your sum} &= \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} \dots \\
\\
{\rm e} &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}\dots
\end{align}$$
A: $$f(x) = x e^{x} = \sum_{n=1}^{\infty} \frac{x^{n}}{(n-1)!} $$
Integrate to get
$$\int_0^x dt \, t e^t =  \sum_{n=1}^{\infty} \frac{x^{n+1}}{(n+1)(n-1)!} $$
Integrate by parts...
$$\int_0^x dt \, t e^t = x e^x - \int_0^x dt \, e^t = (x-1) e^x +1$$
Plug in $x=1$ on both sides to get
$$\sum_{n=1}^{\infty} \frac{1}{(n+1)(n-1)!} = 1$$
A: Change the summation variable to $k = n+1$.
$$\sum_1^\infty \frac{1}{(n+1)!} = \sum_2^\infty \frac{1}{(k)!} = e - \frac{1}{0!} - \frac{1}{1!} = e-2
$$
A: **Hint: **$$\sum_{n=1}^\infty \frac{1}{(n+1)!} = \sum_{n=2}^\infty \frac{1}{n!} \\ = \sum_{n=0}^\infty \frac{1}{n!}-\frac{1}{1!}-\frac{1}{0!}$$
A: $$
\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{\left( {n - 1} \right)!\left( {n + 1} \right)}}}  = \sum\limits_{n = 1}^{ + \infty } {\frac{n}{{\left( {n + 1} \right)!}}}  = \sum\limits_{n = 0}^{ + \infty } {\frac{n}{{\left( {n + 1} \right)!}}}  = 1
$$
A: Consider $$f(x)=\sum^{\infty}_{n=1}\frac{x^{n+1}}{(n-1)!(n+1)}\Rightarrow f'(x)=\sum^{\infty}_{n=1}\frac{x^{n}}{(n-1)!}=x\sum^{\infty}_{n=1}\frac{x^{n-1}}{(n-1)!}=x\sum^{\infty}_{n=0}\frac{x^{n}}{n!}=xe^x$$
Therefore 
$$f(x)=\int xe^x\,dx=xe^x-e^x+c$$
where $c=f(0)+e^0-0\cdot e^0=1$ so at $x=1$ you get your sum 
$$f(1)=\sum^{\infty}_{n=1}\frac{1}{(n-1)!(n+1)}=1\cdot e^1-e^1+c=c=1$$
A: To evaluate the second series note:
$$ e=\sum_{n=0}^\infty\frac{1}{n!}=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots$$
$$ \sum_{n=1}^\infty\frac{1}{(n+1)!}=\frac{1}{2!}+\frac{1}{3!}+\dots$$
Thus the second series is just $e-2$.  If you combine the two series you get $1$.
