<s>[INST] 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 $$\sum_{n=0}^\infty\frac{1}{n!}$$?
 [/INST] $$f(x) = x e^{x} = \sum_{n=1}^{\infty} \frac{x^{n}}{(n-1)!} $$
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
$$\sum_{n=1}^{\infty} \frac{1}{(n+1)(n-1)!} = 1$$
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