# Prove $\sum_{n=1}^\infty(e-\sum_{k=0}^n\frac1{k!})=1$

This comes from the comments section of this question here, credits Lucian.

The statement is

$$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right)=1$$

This looks really interesting, so I was wondering if anyone has any ideas/suggestions or even better the whole proof?

Disclaimer: I added that link specifically because this might look like homework, leading to "What have you tried?". So the disclaimer is, I actually haven't tried anything yet, because yes, asking the community is just easier. It would be great if someone could answer this though since I, personally at least, think that this equation is very interesting and adds value to any math enthusiast stumbling upon it here someday. :) I will myself be trying it meanwhile too though(it really is interesting) , and add and accept my solution if I am successful and there are no answers yet.

I tried wolfram alpha for $\sum_{k=0}^n\frac1{k!}$ and it yields

$$\sum_{k=0}^n\frac1{k!} = \frac{e\Gamma (n+1,1)}{\Gamma(n+1)}$$

where $\Gamma(a,r)$ is the incomplete gamma function and $\Gamma(a)$ is the euler gamma function. Documentation here. Might be of help, might not. Just thought it'd be worth adding.

The first thought that comes into my head is to write $$e - \sum_{k=0}^n \frac{1}{k!} = \sum_{k=n+1}^\infty \frac{1}{k!},$$ so that the given sum is equivalent to a double sum: \begin{align*} \sum_{n=1}^\infty \sum_{k=n+1}^\infty \frac{1}{k!} &= \sum_{k=2}^\infty \sum_{n=2}^k \frac{1}{k!} \\ &= \sum_{k=2}^\infty \frac{1}{(k-2)!k} \\ &= \sum_{k=2}^\infty \frac{1}{(k-1)!} - \frac{1}{k!} = 1. \end{align*}. Of course, a somewhat more rigorous argument is needed to justify the interchange of the order of summation. A minor modification to consider the original series' partial sums, and taking the limit to get the double infinite sum, is sufficient and is left as an exercise.

• Oh that was fast. Thanks. – Guy Mar 7 '14 at 18:11
• All the terms are positive, so it is okay to exchange sums. – abnry Mar 7 '14 at 18:12
• I am over my daily vote limit. I'll upvote this tomorrow. – Guy Mar 7 '14 at 18:17
• @nayrb of course, you are right. ^_^ – heropup Mar 7 '14 at 18:17

$$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right) =\sum_{n=1}^\infty \int_0^1 \exp(u) \frac{(1-u)^{n}}{n!} du$$as everything is positive: $$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right)= \int_0^1 \exp(u) \sum_{n=1}^\infty \frac{(1-u)^{n}}{n!} du \\= \int_0^1 \exp(u)(\exp(1-u) - 1) du = e - \int^1_0 \exp u du = 1$$

• Might be missing something obvious, but $$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right) =\sum_{n=1}^\infty \int_0^1 \exp(u) \frac{(1-u)^{n}}{n!} du$$ how? – Guy Mar 7 '14 at 18:23
• this is the Taylor formula, obtained via $n$ integrations by parts. – mookid Mar 7 '14 at 18:23
• Ah, I am unfamiliar with this. I'll try to prove this myself though. Rest of the solution is great. (+1)(after my vote limit time restriction is lifted that is) – Guy Mar 7 '14 at 18:25
• you obtain it via $\exp 1 - 1 = \int_0^1 1 \exp u du = \ldots$ integrating the polynomial and derivating the $\exp$. – mookid Mar 7 '14 at 18:27
• Whoa dude, spoiler alert. :p – Guy Mar 7 '14 at 18:30