Modified exponential summation

Question

How do we prove that $$1+\frac{2^3}{2!}+\frac{3^3}{3!}+\frac{4^3}{4!}+\cdots=5e$$

I'll post the answer to this question as a knowledge share.

Answer 1

Let $S$ denote the sum in question.

$$S=\sum\limits_{r=1}^{\infty}\left(\frac{r^3}{r!}\right)$$

Here, the general term
$$\begin{equation}\begin{aligned} t_r &= \frac{r^3}{r!}\\ &=\frac{r^2}{(r-1)!}\\ &=\frac{(r-1+1)^2}{(r-1)!}\\ &=\frac{(r-1)^2}{(r-1)!}+\frac{1}{(r-1)!}+\frac{2(r-1)}{(r-1)!}\\ &=\frac{r-1}{(r-2)!}+\frac{1}{(r-1)!}+\frac{2}{(r-2)!}\\ &=\frac{r-2+1}{(r-2)!}+\frac{1}{(r-1)!}+\frac{2}{(r-2)!}\\ &=\frac{r-2}{(r-2)!}+\frac{1}{(r-2)!}+\frac{1}{(r-1)!}+\frac{2}{(r-2)!}\\ &=\frac{1}{(r-3)!}+\frac{3}{(r-2)!}+\frac{1}{(r-1)!}\\ \end{aligned}\end{equation}$$

Splitting $S$ to my convenience, I’ll write:

$$\begin{equation}\begin{aligned} S&=1+\frac{2^3}{2!}+\sum\limits_{r=3}^{\infty}\left(\frac{r^3}{r!}\right)\\ &=1+\frac{2^3}{2!}+\sum\limits_{r=3}^{\infty}\left\{\frac{1}{(r-3)!}+\frac{3}{(r-2)!}+\frac{1}{(r-1)!}\right\}\\ \end{aligned}\end{equation}$$

From the exponential series, we have

$$e^x=\sum\limits_{r=0}^{\infty}\left(\frac{x^r}{r!}\right)$$ $$\implies e=\sum\limits_{r=0}^{\infty}\left(\frac{1}{r!}\right)$$

Utilizing this result, we can write,

$$ S=\left(1+\frac{2^3}{2!}\right)+e+(3e-3)+(e-2)=5e $$