The sum 1 + 1+32! + 1+3+53! + 1+3+5+74! + is equal to:

Question

The sum $1 + \dfrac{1+3}{2!} + \dfrac{1+3+5}{3!} + \dfrac{1+3+5+7}{4!} + \cdots\infty$ is equal to:

Accepted Answer

The $r$-th term is $T_r = \dfrac{r^2}{r!} = \dfrac{r}{(r-1)!} = \dfrac{(r-1)+1}{(r-1)!} = \dfrac{1}{(r-2)!}+\dfrac{1}{(r-1)!}$.

$$\sum_{r=1}^{\infty}T_r = \sum_{r=1}^{\infty}\frac{1}{(r-2)!}+\sum_{r=1}^{\infty}\frac{1}{(r-1)!} = e+e = 2e.$$

Solutions · 1