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I read that $e = \sum_{i=0}^\infty$$ 1\over n!$. This isn't immediately obvious to me, and I can't find proof of this. Can somebody explain to me, how do I prove this from definition $e = \lim_{n\to \infty}$$ (1 + {1\over n})^n $, or point me to one?

Anixx
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Dark Archon
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3 Answers3

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If you expand $\left(1+\frac1n\right)^n$, you get $$\tag1 \sum_{k=0}^n{n\choose k}\frac1{n^k}=\sum_{k=0}^n\frac{n!}{(n-k)!n^k}\cdot\frac1{k!}$$ The factor $\frac{n!}{n^k(n-k)!}$ is $\le 1$ (the denominator is the numerator with some factors $n-i$ increased to $n$), hence the expression in $(1)$ is always $<\sum_{k=0}^\infty\frac1{k!}$, hence $\mathrm e\le\sum_{k=0}^\infty\frac1{k!}$.

On the other hand, For any given $k$, we have $\frac{n!}{(n-k)!n^k}\to1$ as $n\to\infty$. Hence for any $\epsilon>0$ and any $m\in\mathbb N$, we can pick $n$ so large that the first $m$ summands in $(1)$ exceed $\sum_{k=0}^m\frac{1-\epsilon}{k!}$. As all summands are positive, we conclude that $\left(1+\frac1n\right)^n>(1-\epsilon)\sum_{k=0}^m\frac1{k!} $, hence $(1-\epsilon)\sum_{k=0}^\infty\frac1{k!}\le\mathrm e$ for all $\epsilon>0$.

Hagen von Eitzen
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Expand using Binomial theorem: $$ \begin{align} \lim_{N\rightarrow\infty}(1 + \frac{1}{N})^N &= \lim_{N\rightarrow\infty} \sum\limits_{n=0}^N {N \choose n} \left(\frac{1}{N}\right)^n \\ &= \lim_{N\rightarrow\infty} \sum\limits_{n=0}^N \frac{N!}{n! (N-n)! N^n} \\ &= \lim_{N\rightarrow\infty} \sum\limits_{n=0}^N \frac{N \times (N-1)\times (N-2) \times \cdots \times (N-n+1)}{n! N^n} \\ &= \lim_{N\rightarrow\infty} \sum\limits_{n=0}^N \frac{N^n}{n! N^n} \\ &= \sum\limits_{n=0}^\infty \frac{1}{n!} \end{align} $$

EDIT: I like Hagen von Eitzen's explanation better than my own. Going from the 3rd to 4th line I said that each factor in the numerator approaches $N$ as $N \rightarrow \infty$ but glossed over the fact that $n$ gets summed to $N$, so that step is not as obvious as I made it out to be.

Bridgeburners
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  • To make that step you mention more intuitive: Pick any ratio $r$ between 0 and 1. There are only finitely man $n$ for wich $\frac{n}{N} \leq r$. But since $N$ goes to infinity, the proportion of the fraction (pairs of $N- n$ and $N$) that is smaller than $r$ gets increasingly small and so the average proportion of pairs of $N- n$ and $N$ tends to $1$ with rising $N$. – user3578468 Jun 16 '20 at 05:38
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This isn't exactly a "proof," but just to give you an idea: let $f(x)=e^x$. Then $f'(x)=e^x$ and hence $f^{(n)}(x)=e^x$ for $n=1,2,3,\ldots$. So $f^{(n)}(0) = 1$ for all $n$. Then the Taylor series for $e^x$ centered at zero is

$$ \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^\infty \frac{x^n}{n!} $$

Plugging in $x=1$ we see that $e = \sum_{n=0}^\infty \frac1{n!}$.

Math1000
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