This question is from Courant Calculus/Analysis book Vol 1, Section 1.7 Question 7:
Show that $ \frac{1}{e}=1-1+\frac{1}{2!}-\frac{1}{3!}+...+\frac{(-1)^n}{n!}+... $. Hint: Consider the product of the nth partial sums of the expansions for e and 1/e.
It seems that after doing partial sums for $e$ and the supposed $1/e$ we should have $1+c$ for some error value $c$. Then we take limits to infinity and $c$ should be $0$. I find the products of those sums too confusing and don't know how to get $1+c$.
Define $e_n = \sum_{i = 0}^n \frac1{i!}$ and $f_n = \sum_{j = 0}^n\frac{(-1)^j}{j!}$. The hope is that $\lim_{n\to \infty} e_nf_n = 1$. Is that what really happens? $$ \lim_{n\to \infty}e_nf_n = \lim_{n\to \infty}\left(\sum_{i = 0}^n \frac1{i!}\right)\left(\sum_{j = 0}^n\frac{(-1)^j}{j!}\right)\\ = \lim_{n\to \infty}\sum_{i = 0}^n \left(\frac1{i!}\cdot \sum_{j = 0}^n\frac{(-1)^j}{j!}\right)\\ = \lim_{n\to \infty}\sum_{i, j = 0}^n\frac{(-1)^j}{i!\cdot j!} $$ At this point, I wish to split the sum up into two parts: One part for when $i+j \leq n$, and one for $i+j > n$. I'll handle the second part first.
There are $n(n+1)/2$ terms in this second part. The largest term (in absolute value) is strictly smaller than $\frac{1}{(n/2)!^2}$ (if you'll excuse my use of factorial on something which is possibly not an integer). The exact size is $$ \text{Largest term in the second part } = \cases{\dfrac{1}{\frac{n}{2}!\cdot \frac{n+2}{2}!} & if $n$ is even\\ \dfrac{1}{\left(\frac{n+1}{2}!\right)^2}& if $n$ is odd} $$ and can be used in the argument to below, but for typographical and reading simplicity I'll leave the slightly invalid (because of non-integers) approximation in place. The sum of all of the terms in the second part is therefore less than $$ \frac{n(n+1)}{(n/2)!^2} $$ which we can see goes toward $0$ as $n$ increases, for instance by noting that $n(n+1)< 2\cdot 3 \cdot \frac n2\cdot \frac n2$, and what's left in the denominator is at least $\frac n2 - 1$.
So we're left with the first part. That is, $$ \sum_{i+j \leq n\\i, j\geq 0}\frac{(-1)^j}{i!j!} $$ Let's take this one diagonal at a time. In other words, decompose this sum into $$ \sum_{k = 0}^n\sum_{i + j = k\\i, j \geq 0}\frac{(-1)^j}{i!j!} $$ Here we can see that the inner sum is $1/k!$ times the binomial expansion of $(1 - 1)^k$ (at least for $k > 0$; I won't get into $0^0$ discussions here). So the sum of the first part is exactly equal to $1$.
This concludes the proof.