Find the value of $\displaystyle \sum_{m=1}^\infty \sum^{\infty}_{n=1}\frac{m\cdot n}{(m+n)!}$.
My try: $$\sum^{\infty}_{m=1}\bigg(\frac{m}{(m+1)! }+\frac{2m}{(m+2)!}+\cdots\cdots +\frac{m+\infty}{(m+\infty)!}\bigg)$$
I did not understand how to start it.
I could use some help. Thanks.
Best Answer
Note that by letting $k=m+n$, we have that (terms can be arranged because they are non-negative), $$\begin{align} \sum^{\infty}_{m=1}\sum^{\infty}_{n=1}\frac{m\cdot n}{(m+n)!} &=\sum^{\infty}_{k=2}\frac{1}{k!}\sum_{n=1}^{k-1}n(k-n)= \frac{1}{6}\sum^{\infty}_{k=2}\frac{k^3-k}{k!}\\&=\frac{1}{6}\sum^{\infty}_{k=2}\frac{k(k-1)(k-2)+3k(k-1)}{k!}\\ &= \frac{1}{6}\sum^{\infty}_{k=3}\frac{1}{(k-3)!}+\frac{1}{2}\sum^{\infty}_{k=2}\frac{1}{(k-2)!}\\ &=\frac{e}{6}+\frac{e}{2}=\frac{2e}{3}.\end{align}$$