Proof of the Series Representation of e...stuck

Question

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!

Answer 1

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.

Answer 2

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*}

Answer 3

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 $x=1$.