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?

  • $\begingroup$ In your last line, the first term should be $(e-1)$, not $e$. Note that:$$e=\frac1{0!}+\frac1{1!}+\frac1{2!}+ \dotsb\ne\frac1{1!}+\frac1{2!}+\dotsb$$ $\endgroup$ – Akiva Weinberger Jan 14 '16 at 1:04
  • $\begingroup$ @AkivaWeinberger You're right, slight typo. Corrected. $\endgroup$ – user170231 Jan 14 '16 at 3:03

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.

  • 3
    $\begingroup$ How you reverse the order could be explained in more detail - it's probably something the person asking hasn't encountered much, and is not quite so trivial. $\endgroup$ – vrugtehagel Jan 13 '16 at 23:36
  • $\begingroup$ it is trivial that if $\sum_k \sum_n |a_{n,k}|$ converges then $\sum_n \sum_k |a_{n,k}|$ also converges and in any order it will still converge : this is the absolute convergence theorem. finally, proving that they converge to the same limit is a matter of proving the residual tends to $0$ (still the absolute convergence theorem) $\endgroup$ – reuns Jan 13 '16 at 23:41
  • $\begingroup$ @vrugtehagel I expanded the answer to hopefully give some intuition about that. $\endgroup$ – user296602 Jan 13 '16 at 23:46
  • $\begingroup$ To $ $explain the reversing of order (the colons mean "such that"):\begin{align}\sum_{k=1}^\infty\sum_{n=1}^k&=\sum_{k:1\le k}\ \sum_{n:1\le n\le k}\\&=\sum_{k,n:1\le k\text{ and }1\le n\le k}\\&=\sum_{k,n:1\le n\le k}\\&=\sum_{k:n\le k}\ \sum_{n:1\le n\le k}\\&=\sum_{k=1}^\infty\sum_{n=1}^k\end{align}I probably should've used the Iverson bracket. It would've been a lot clearer. $\endgroup$ – Akiva Weinberger Jan 14 '16 at 1:09
  • 2
    $\begingroup$ @user170231 I wouldn't really call this telescoping, but notice that if you expand $e$ in each of your braces, the first one contributes $1/2, 1/6, 1/24, ...$ each once. The next brace contributes $1/6, 1/24, ...$ each once. The next brace contributes $1/24, ...$ each once. The number of occurrences of $1/2$ is then $1$, the number of occurrences of $1/6$ is $2$, and so on - simplifying this gives a way to sum the series (which is essentially the same as reversing the order). $\endgroup$ – user296602 Jan 14 '16 at 22:47

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