Is there a direct way 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 definitions of $e$ directly?

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

In other words, I want to know a proof that is not "prove $e^x=\sum\limits_{n=0}^\infty\frac{1}{n!}x^n$ and then substitute $x=1$". I guess we do not need $e^x$ to prove this equality.

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 $\sum\limits_{k=1}^\infty\lim\limits_{n\to\infty}\frac{1}{n}=0$.

None of the answers to Prove the definitions of e to be equivalent answers this question. There are 3 answers to it:

1 is begging the question. It assumes $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$.

2 is wrong. The operation of limits is incorrect.

3 is not direct proof.

I did it there https://math.stackexchange.com/a/2172435/399263 use binomial expansion then dominated convergence theorem to prove it is equal to the series. Yet there are plenty of other answers on the same subject on MSE. This one for instance https://math.stackexchange.com/q/2821095/399263 — zwim, Oct 06 '20 at 10:39
@Aryadeva Pleas read the question carefully. This is not repost. — , Oct 06 '20 at 10:40
If you want a detailed explanation of a particular step in those proofs, I suggest you make this more explicit. — Arnaud D., Oct 06 '20 at 10:41
@ArnaudD. They are wrong proofs using incorrect operation of limits. I don't want detail of them. — , Oct 06 '20 at 10:51
Switching the limit and summation is justified: see Under what condition we can interchange order of a limit and a summation? — Toby Mak, Oct 06 '20 at 10:52
There are many questions about this, if this one does not satisfy you then you can try looking at the linked questions. — Arnaud D., Oct 06 '20 at 10:56
@PONPON The dominated convergence theorem applies to only certain types of functions, and one of them is $f(x) = e^x$. — Toby Mak, Oct 06 '20 at 10:58
@PONPON What do you gain from excluding any other definitions of $e$ from the proof? These two definitions have been arbitrarily chosen, compared to the many definitions for $e$ that exist. — Toby Mak, Oct 06 '20 at 11:02
@TobyMak If what you write is true, it needs proof. So the answers are inadequate. I think they do not know that theorem and do wrong opetation. — , Oct 06 '20 at 11:02
@TobyMak I want to know direct proofs of equality between all pairs of definitions of $e$ but they are too large to one question, so I ask one. — , Oct 06 '20 at 11:06
Look also at Given that $\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e$, show that $1+\frac{1}{1!}+\frac{1}{2!}+\cdots=e$. — Sil, Oct 06 '20 at 11:50

SomeStrangeUser · Answer 1 · 2020-10-06T12:30:36.367

A nice direct proof comes from Rudin:

Let $$ s_n=\sum_{k=0}^n\frac{1}{k!}\;,\;\;t_n=\left(1+\frac{1}{n}\right)^{n} $$ First, $s_n$ converges, as it is an increasing sequence that is bounded above by $1 + \sum_{k=0}^{n-1}\frac{1}{2^k} \le 3$.

Let $e:=\lim_{n\rightarrow\infty}s_n$. By the binomial theorem, $$ \small{t_n = 1 + 1+\frac{1}{2!}\left(1-\frac{1}{n}\right) + \frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) +\; \cdots \;+ \frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{n-1}{n}\right)} $$ Hence, $t_n\le s_n$, so that $$\limsup_{n\rightarrow\infty}t_n\le e$$

Next, if $n\ge m$, $$ t_n\ge 1 + 1 + \frac{1}{2!}\left(1-\frac{1}{n}\right)+ \cdots + \frac{1}{m!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{m-1}{n}\right) $$ Let $n\rightarrow\infty$, keeping $m$ fixed. We get $$ \liminf_{n\rightarrow\infty}t_n\ge 1+1+\frac{1}{2!} + \cdots + \frac{1}{m!} $$ so that $s_m\le\liminf_{n\rightarrow\infty} t_n$. Letting $m\rightarrow\infty$, we finally get: $$ e\le\liminf_{n\rightarrow\infty}t_n $$

freakish · Accepted Answer · 2020-10-06T11:29:26.997

0

In the original series you can enter with $\lim$ under summation because it is dominated by $\sum_{k=1}^\infty\frac{1}{k!}$ which is convergent. Read more about domination here: https://en.wikipedia.org/wiki/Dominated_convergence_theorem (and the simplification for series: Discrete version of dominated convergence thm ).

But yes, this is not as trivial as other answers present. I'm not aware of an easier solution.

edited Oct 06 '20 at 11:29

answered Oct 06 '20 at 11:17

freakish

42,851

@PONPON I've updated the answer. – freakish Oct 06 '20 at 11:26
This approach is given as an example in Wikipedia. – metamorphy Oct 06 '20 at 12:26

Is there a direct way to prove $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n=\sum\limits_{n=0}\frac{1}{n!}$?

2 Answers2