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Bottom Line: Prove that $e = 1+1+\frac{1}{2!}+\cdots$

Define $e = \lim\limits_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n$

I would like to do it by expanding $\left(1+\frac{1}{n}\right)^n$ binomially as $\sum\limits_{j=0}^n {n \choose j}\left(\frac{1}{n}\right)^j$. Somehow I need to show that this approaches $\frac{1}{j!}$ as $n$ approaches $\infty$. That's basically it I think. I feel like I'm missing something easy, but I haven't been able to find this online except for a Dr. Math forum that I can't follow (here: Link). Please be thorough in your explanation, if possible.

Thanks for the help!

Glorfindel
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daniel.wright
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3 Answers3

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If you want a maximum level of rigor, you can avoid the interchange of two limiting operator problem as follows:

As we observed,

$$ \left( 1 + \frac{1}{n} \right)^n = \sum_{j=0}^{n} \binom{n}{j}\frac{1}{n^j} = \sum_{j=0}^{n} \frac{1}{j!}\prod_{k=1}^{j} \left( 1 - \frac{k-1}{n} \right) $$

where the product is considered to yield $1$ if $j = 0$ by convention. Now, fix a positive integer $m$. Then by noting that each summand is non-negative, for any $n \geq m$ we have

\begin{align*} \sum_{j=0}^{m} \frac{1}{j!}\prod_{k=1}^{j} \left( 1 - \frac{k-1}{n} \right) &\leq \sum_{j=0}^{n} \frac{1}{j!}\prod_{k=1}^{j} \left( 1 - \frac{k-1}{n} \right) \leq \sum_{j=0}^{n} \frac{1}{j!}. \end{align*}

This inequality shows that

\begin{align*} \liminf_{n\to\infty} \sum_{j=0}^{m} \frac{1}{j!}\prod_{k=1}^{j} \left( 1 - \frac{k-1}{n} \right) &\leq \liminf_{n\to\infty} \left(1 + \frac{1}{n}\right)^{n} \\ &\leq \limsup_{n\to\infty} \left(1 + \frac{1}{n}\right)^{n} \leq \limsup_{n\to\infty} \sum_{j=0}^{n} \frac{1}{j!}. \end{align*}

But since

$$ \liminf_{n\to\infty} \sum_{j=0}^{m} \frac{1}{j!}\prod_{k=1}^{j} \left( 1 - \frac{k-1}{n} \right) = \sum_{j=0}^{m} \frac{1}{j!} $$

and

$$ \limsup_{n\to\infty} \sum_{j=0}^{n} \frac{1}{j!} = \sum_{j=0}^{\infty} \frac{1}{j!}, $$

it follows that

$$ \sum_{j=0}^{m} \frac{1}{j!} \leq \liminf_{n\to\infty} \left(1 + \frac{1}{n}\right)^{n} \leq \limsup_{n\to\infty} \left(1 + \frac{1}{n}\right)^{n} \leq \sum_{j=0}^{\infty} \frac{1}{j!}. $$

Notice that this inequality holds for any $m$. Thus taking $m\to\infty$, we obtain the desired result.

Sangchul Lee
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  • +1 (if old!) happened to be looking for a forgotten clean rigorous proof (with no mention of l'hopital's / log etc), dimly remembered it involved appropriate liminf/limsup specification, nice to find a delightfully clear and succinct exposition thanks! – Mehness Oct 04 '18 at 13:07
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As you point out, the desired formula follows without too much trouble if you can show that $$ \lim_{n\to\infty} \binom nj \bigg(\frac{1}{n}\bigg)^j = \frac{1}{j!}. $$ To see this, note that \begin{align*} \binom nj \bigg(\frac{1}{n}\bigg)^j &= \frac{n(n-1)(n-2)\cdots(n-(j-1))}{j!} \bigg(\frac{1}{n}\bigg)^j \\\ &= \frac{n(n-1)(n-2)\cdots(n-(j-1))}{n^j} \frac{1}{j!} \\\ &= 1\bigg(1-\frac1n\bigg)\bigg(1-\frac2n\bigg)\cdots\bigg(1-\frac{j-1}n\bigg) \frac{1}{j!}. \end{align*} Therefore \begin{align*} \lim_{n\to\infty} \binom nj \bigg(\frac{1}{n}\bigg)^j &= \lim_{n\to\infty} 1\bigg(1-\frac1n\bigg)\bigg(1-\frac2n\bigg)\cdots\bigg(1-\frac{j-1}n\bigg) \frac{1}{j!} \\\ &= \lim_{n\to\infty} \bigg(1-\frac1n\bigg) \lim_{n\to\infty} \bigg(1-\frac2n\bigg) \cdots \lim_{n\to\infty} \bigg(1-\frac{j-1}n\bigg) \cdot \frac1{j!} \\\ &= 1\cdot 1\cdots 1\cdot \frac1{j!} = \frac1{j!}. \end{align*}

Greg Martin
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    You need to justify why $\sum_{j=1}^n a_{n,j}\to \sum_{j=1}^\infty a_{\infty,j}$ which is not always true. – Pedro Mar 17 '13 at 03:23
  • I believe it is enough that the $a_{n,j}$ are positive and increasing? – Alex Zorn Mar 17 '13 at 03:52
  • I'm not sure what Peter's hinting at, but the step one before the last is definitely wrong: one can NOT use arithmetic of limits when the terms of the sum (or multiplication) depend on the variable going to infinity, as in this case. One needs some kind of bounding and then taking the limit. – DonAntonio Mar 17 '13 at 04:51
  • Yep, you're all right (that's why I said "without too much trouble"). @DonAntonio , you can view each $\sum_j \binom nj n^{-j}$ as an infinite sum, since $\binom nj=0$ when $n<j$. Then the fact that they're positive and increasing does do the trick (Dominated Convergence Theorem). – Greg Martin Mar 17 '13 at 07:39
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This might be cheating, but use: $$f(x)=e^x = \lim\limits_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^{nx}$$ So that: $$f^{(k)}(x) = \lim\limits_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^{nx} \left(\log\left(1+\frac{1}{n}\right)^n\right)^k = e^x$$

Now write the taylor series for $e^x$ using the above and plug in $x=1$.

Nathaniel Bubis
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