Question

Let $S_n=1\cdot (n-1)+2\cdot (n-2)+3\cdot (n-3)+\cdots +(n-1)\cdot 1,n\ge 4.$ The sum $\sum _{n=4}^{\infty }(\frac{2S_n}{n!}-\frac{1}{(n-2)!})$ is equal to

• A. $\frac{e-1}{3}$
• B. $\frac{e}{3}$
• C. $\frac{e}{6}$
• D. $\frac{e-2}{6}$

Answer 1

The correct option is A $\frac{e-1}{3}$

$S_n=1\cdot (n-1)+2\cdot (n-2)+3\cdot (n-3)+\cdots +(n-1)\cdot 1,n\ge 4.$
i.e. $T_k=k(n-k)$

$S_n=\sum _{k=1}^n{T}_k=\sum _{k=1}^n(kn-k^2)$
$=\frac{n\left(n\right(n+1\left)\right)}{2}-\frac{n(n+1)(2n+1)}{6}$
$=\frac{n(n+1)}{2}\left(\frac{3n-(2n+1)}{3}\right)=\frac{n(n^2-1)}{6}=S_n$

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

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