$\text{ Let }{S}_n=1·(n-1)+2·(n-2)+3·(n-3)+\dots +(n-1)·1, n⩾4.$ 

$\text{The sum }\sum _{\mathrm{n}=4}^{\infty }\left(\frac{2{ \mathrm{S}}_{\mathrm{n}}}{\mathrm{n}!}-\frac{1}{(\mathrm{n}-2)!}\right)\text{ is equal to :}$ 

$$\begin{gathered} \text{ Let }{S}_n=1·(n-1)+2·(n-2)+3·(n-3)+\dots +(n-1)·1, n⩾4. \\ \mathrm{General} \mathrm{form} \mathrm{of} {\mathrm{S}}_{\mathrm{n}}=\sum _{r=1}^{n-1}r(n-r)=\frac{n\left(n^2-1\right)}{6} \end{gathered}$$

$\frac{2{ \mathrm{S}}_n}{n!}=\frac{2n\left(n+1\right)\left(n-1\right)}{6n\left(n-1\right)\left(n-2\right)!}=\frac{(n+1)}{3(n-2)!}$

$\Rightarrow \sum _{n=4}^{\infty }\left(\frac{2S_n}{n!}-\frac{1}{(n-2)!}\right)$$\Rightarrow \sum _{n=4}^{\infty }\left(\frac{(n+1)}{3(n-2)!}-\frac{1}{(n-2)!}\right)$

$=\sum _{n=4}^{\infty }\frac{(n-2)}{3(n-2)!}$

$=\frac{1}{3}\sum _{n=4}^{\infty }\frac{1}{(n-3)!}$

$=\frac{1}{3}\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots ..\right)=\frac{e-1}{3}$