<s>[INST] Value of $1/e$ with the partial sum of $e$ How to show $$1/e=1-1+\frac1{2!}-\frac1{3!}+\frac1{4!}+\dots$$
I have consider the product of the nth partial sums of the expansions for $e$ and $1/e$ but still can't prove there are equal to 1
with the definition $e=1+1+\frac1{2!}+\frac1{3!}+\dots$
 [/INST] I assume you know that 
$$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots = \sum_{k=0}^{\infty} \frac{1}{k!}.$$
Now let 
$$f = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \ldots = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!}.$$
By the Mertens theorem and the binomial formula
$$\begin{align*}
e \cdot f & = \left( \sum_{k=0}^{\infty} \frac{1}{k!} \right) \cdot \left( \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \right) = \sum_{n=0}^{\infty} \sum_{k=0}^n \frac{(-1)^{n-k}}{k! (n-k)!} \\[1ex]
& = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^n \binom{n}{k} \cdot 1^k \cdot (-1)^{n-k} = \sum_{n=0}^{\infty} \frac{(1-1)^n}{n!} = \frac{1}{0!} = 1 
\end{align*}$$
so $f = \frac{1}{e}$. In the end we used the fact that 
$$0^n = \begin{cases} 1 & \text{ if } n = 0 \\ 0 & \text{ if } n > 0 \end{cases}$$
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