I can't seem to come to grips with the result below:
$$S=\sum_{n=1}^\infty \sum_{k=n}^\infty\frac{1}{k!}=e$$
which is given by Mathematica (code below) and (numerically) verified by WolframAlpha.
In[65]:= Sum[1/k!, {n, 1, Infinity}, {k, n, Infinity}]
Out[65]= E
I've attempted to work it out in the following way:
$$\begin{align*}S&=\sum_{n=1}^\infty\sum_{k=n}^\infty \frac{1}{k!}\\[1ex]
&=\sum_{n=1}^\infty\left(\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\cdots\right)\\[1ex]
&=\sum_{n=1}^\infty\frac{1}{n!}+\sum_{n=1}^\infty\frac{1}{(n+1)!}+\sum_{n=1}^\infty\frac{1}{(n+2)!}+\cdots\\[1ex]
&=\sum_{n=1}^\infty\frac{1}{n!}+\sum_{n=2}^\infty\frac{1}{n!}+\sum_{n=3}^\infty\frac{1}{n!}+\cdots\\[1ex]
&=(e-1)+\left(e-1-\frac{1}{2}\right)+\left(e-1-\frac{1}{2}-\frac{1}{6}\right)+\cdots\end{align*}$$
which doesn't appear to me to follow a telescoping pattern, but I might be wrong about that. It's not obvious to me if this actually does telescope.
Edit: Changing the order of summation does wonders, as shown in the accepted answer, but I'm currently wondering if there is any possibility that the last line admits any neat telescoping argument?
Answer
Reverse the order of summation and this becomes
\begin{align*}
\sum_{k = 1}^{\infty} \sum_{n = 1}^k \frac{1}{k!} &= \sum_{k = 1}^{\infty} \frac 1 {k!} \sum_{n = 1}^k 1\\
&= \sum_{k = 1}^{\infty} \frac{1}{k!} \cdot k \\
&= \sum_{k = 1}^{\infty} \frac{1}{(k - 1)!} = e
\end{align*}
To understand the change of order, note that all sums here are very convergent (and positive), so I'm not going to worry about technical issues. The original sum is about fixing $n$ and summing over $k \ge n$. If you imagine writing out all the pairs of natural numbers in a grid with $k$ running horizontally and $n$ vertically, this is fixing a column and adding up every pair below the main diagonal. That is, the lower left half of the grid.
On the other hand, we can also describe this as summing over every row, but stopping when we get to the main diagonal.
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