Modified exponential summation 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.
 A: 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,
$$
S=\left(1+\frac{2^3}{2!}\right)+e+(3e-3)+(e-2)=5e
$$
