calculus - Easy Double Sums Question: $sum_{m=1}^{infty} sum_{n=1}^{infty} frac{1}{(m+n)!}$

How to calculate $\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{(m+n)!} $ ?

I don't know how to approach it . Please help :)

P.S.I am new to Double Sums and am not able to find any good sources to study it , can anyone help please ?

Answer

This is the same as

$$
\sum_{m=1}^\infty \sum_{k=m+1}^\infty \frac{1}{k!}
$$
We can rearrange terms, noting that for each value of $k$ there will be terms only with
$k > m$. There are $k-1$ possible values of $m$ that satisfy $k>m$. So
$$
\sum_{m=1}^\infty \sum_{k=m+1}^\infty \frac{1}{k!} = \sum_{k=1}^\infty \frac{k-1}{k!}
$$
The last trick is to note that it will be much easier to sum $\frac{k}{k!}$ so break up the numerator:
$$

\sum_{k=1}^\infty \frac{k-1}{k!} = \sum_{k=1}^\infty \frac{k}{k!} - \sum_{k=1}^\infty \frac{1}{k!} = \sum_{k=1}^\infty \frac{1}{(k-1)!}- \sum_{k=1}^\infty \frac{1}{k!}
$$
And this in turn is
$$
\frac{1}{0!} + \sum_{k=2}^\infty \frac{1}{(k-1)!} - \sum_{k=1}^\infty \frac{1}{k!}
= 1 +\sum_{j=1}^\infty \frac{1}{j!} - \sum_{k=1}^\infty \frac{1}{k!}
$$
So far, only rearrangement of terms has happened. Now we note that $\sum_{k=1}^\infty \frac{1}{k!}$ is absolutely convergent, so the rearrangement of terms is valid; and the tow sums left cancel, so the answer is
$$1
$$