Evaluate $S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}.$ 
Prove that $$S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\frac{e}{2}-1.$$

$$
S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}\\
=\frac{1}{2}\bigg[\frac{1}{3.0!}+\frac{1}{4.1!}+\frac{1}{5.2!}+\frac{1}{6.3!}+\dots\bigg]=\frac{1}{2}\bigg[\Big(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots\Big)-\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big)\bigg]\\=\frac{e}{2}-\frac{1}{2}\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big)
$$
How do I proceed further as I am stuck with the last infinite series.
 A: Hint: Take two derivation of both side of
$$\dfrac{e^x-1-x-\dfrac12x^2}{x}=\sum_{n=2}^{\infty}\dfrac{x^n}{(n+1)!}$$
then let $x=1$.
A: $$\dfrac{\binom n2}{(n+1)!}=\dfrac{n(n-1)}{2(n+1)!}$$
As $n(n-1)=(n+1)n-2(n+1)+2$
$$ \dfrac{\binom n2}{(n+1)!}=\dfrac12\cdot\dfrac1{(n-1)!}-\left(\underbrace{\dfrac1{n!}-\dfrac1{(n+1)!}}\right)$$
The terms under brace telescopes
Use $e^y=\displaystyle\sum_{r=0}^\infty\dfrac{y^r}{r!}$ for the first term
See also : Evaluate the series $\lim\limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n+2}{2(n-1)!}$
A: $$S=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}=\sum_{n=0}^\infty\frac{1}{2(n+3)(n)!}$$
$$=\frac{1}{2}\sum_{n=0}^\infty\int_{0}^{1}\frac{x^{n+2}}{n!}dx=\frac{1}{2}\int_{0}^{1}{x^2e^{x}}dx=\frac{e}{2}-1$$
A: First, it must be:
$$S=\sum_{n=2}^\infty\frac{^nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}=\\
\color{blue}{=\sum_{n=2}^\infty\frac{(n+1)-n}{2(n+1)(n-2)!}=\frac12\left[\sum_{n=2}^\infty\frac{1}{(n-2)!}-\sum_{n=2}^\infty\frac{n}{(n+1)(n-2)!}\right]=}\\
=\frac{1}{2}\bigg[\Big(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots\Big)-\Big(\frac{2}{3.0!}+\frac{\color{red}3}{4.1!}+\dots\Big)\bigg]\\=\frac{e}{2}-\frac{1}{2}\Big(\frac{2}{3.0!}+\frac{\color{red}3}{4.1!}+\dots\Big).$$
Second, evaluating the second series is more difficult:
$$\sum_{n=2}^\infty\frac{n}{(n+1)(n-2)!}=\sum_{n=2}^\infty\frac{n^2(n-1)}{(n+1)!}=\\
\sum_{n=2}^\infty\frac{(n+1)(n^2-2n+2)-2}{(n+1)!}=\sum_{n=2}^\infty\frac{(n+1)n(n-1)-(n+1)n+2(n+1)-2}{(n+1)!}=\\
\sum_{n=2}^\infty\left[\frac{1}{(n-2)!}-\frac{1}{(n-1)!}\right]+2\sum_{n=2}^\infty\left[\frac{1}{n!}-\frac{1}{(n+1)!}\right]=\\
\left[\frac1{0!}-\frac1{1!}+\frac1{1!}-\frac1{2!}+\frac1{2!}-\frac1{3!}+\cdots\right]+2\left[\frac1{2!}-\frac1{3!}+\frac1{3!}-\frac1{4!}+\frac1{4!}-\frac1{5!}+\cdots\right]=2.$$
Thus, following lab bhattacharjee's method is more efficient.
