Let S_n = 1(n-1)+2(n-2)+3(n-3)++(n-1)1, n4.

Question

Let $S_n = 1\cdot(n-1)+2\cdot(n-2)+3\cdot(n-3)+\cdots+(n-1)\cdot1$, $n\ge4$. The sum $\displaystyle\sum_{n=4}^{\infty}\left[\frac{2S_n}{n!}-\frac{1}{(n-2)!}\right]$ is equal to:

Accepted Answer

$S_n = \displaystyle\sum_{r=1}^{n-1}r(n-r) = \dfrac{n(n^2-1)}{6} = \dfrac{n(n-1)(n+1)}{6}$.

$$\frac{2S_n}{n!} = \frac{(n+1)}{3(n-2)!} \implies \frac{2S_n}{n!}-\frac{1}{(n-2)!} = \frac{n-2}{3(n-2)!} = \frac{1}{3(n-3)!}.$$

$$\sum_{n=4}^{\infty}\frac{1}{3(n-3)!} = \frac13\left(\frac{1}{1!}+\frac{1}{2!}+\cdots\right) = \frac{e-1}{3}.$$

Solutions · 1