How to prove $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n=\sum\limits_{n=0}\frac{1}{n!}$?

Question

How to prove the equality of the two defintions of $e$?

$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\sum_{n=0}\frac{1}{n!}$$

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There are answers in other questions like this:

$$\begin{aligned} \lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n &= \lim_{n\to\infty}\sum_{k=0}^n \frac{1}{k!} \prod_{m=0}^{k-1}\left(1-\frac{m}{n}\right)\\ &=^?\sum_{k=0}^\infty\left[\frac{1}{k!}\lim_{n\to\infty}\prod_{m=0}^{k-1}\left(1-\frac{m}{n}\right)\right]\\ &=\sum_{k=0}^\infty\frac{1}{k!} \end{aligned}$$

However, I think this operation of limits is wrong. For example, $\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\frac{1}{n}=1$ and $\sum\limits_{k=1}^\infty\lim\limits_{n\to\infty}\frac{1}{n}=0$.

Answer 1

I do not know a direct way to prove this equality. Here is an undirect way.

First, prove that the series $\sum \dfrac{x^n}{n!}$ is uniformly convergent on all compact sets. Then, you can show this series is differentiable and satisfies the differential equation $y' = y$ with initial datum $y(0) = 1$. Thus, this series is equal to the exponential function, and the value at $1$ is by definition $e$. Then $e = \sum_{n\geqslant 0} \dfrac{1}{n!}$.

Secondly, with some analysis \begin{align} \left(1+\frac{1}{n}\right)^n & = \exp \left(n \ln\left(1+\frac{1}{n}\right) \right)\\ &=\exp\left(n\left(\frac{1}{n} + o\left(\frac{1}{n}\right)\right) \right)\\ &= \exp(1 + o(1)) \end{align} and by continuity of $\exp$, the limit is $\exp(1)$ that is by dfinition, $e$.

Finally, we have proved that $\sum \frac{1}{n!} = e = \lim \left(1+\frac{1}{n}\right)^n$.

Answer 2

Consider this expression:

$$ (1 + \frac{1}{n})^j$$

We can write this as:

$$ (1 + \frac{1}{n})^j = 1 + \frac{j}{n} + \frac{j (j-1)}{2!n^2}... $$

Sub, $ j =n$

$$ (1 + \frac{1}{n})^n = 1 + \frac{n}{n} + \frac{n (n-1)}{2!n^2}... $$

Now, take the limit on both sides as $ n \to \infty$

$$ \lim_{n \to \infty} (1 + \frac{1}{n})^n = 1 + 1 + \frac{1}{2!} + \frac{1}{3!}..$$

Or,

$$ \lim_{n \to \infty} (1 + \frac{1}{n})^n = \sum_{k=0}^{\infty} \frac{1}{k!}$$