How do I prove $\exp(1)=e$ if $\exp$ is defined in a series expansion?

Question

Right now I'm on a journey to clear up my confusion about exponential functions. Thanks to your help in this stack, I was able to derive the basic properties of the ln function from the "position of defining ln in terms of integrals".

In this stack, the definitions of exp and e are as follows;

• $$\exp(x):= 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots,\ $$
• $$e:=\lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^{n} , n\in \mathbb{N}.$$

When defined in this way, the relationship between $\exp(x)$ and $e$ seems to be non-trivial . We further assume that;

• The theory on limit operations such as the scissors-out theorem and the theory on convergence conditions for power series are assumed to exist separately and independently.
• For any integer $a>0$ where $n$ is an integer, $a^n$ is assumed to be defined in an inductive way.

My question.

Under these assumptions, how can we mathematically prove the following? $$\exp(1)=e$$

There is a similar stack, but it does not seem to go into the definition of Napier-number. If possible, we focus my questions on the following "How to evaluate from the lower side.

As far as I tried, I was probably able to suppress it from the upper side. In other words, I believe I was able to prove the following.

$$e\le \exp(1)$$

Therefore, perhaps the question is how to find the series to suppress from below.

This means that the most desirable answer is to show $$e\ge \exp(1)$$

Proof of $e\le exp(1)$
From the binomial theorem,

$$\left( 1+\frac{1}{n} \right)^{n} =\sum_{k=0}^{n} {}_{n} \mathrm{C}_{k} \frac{1}{n^k}\\ =\sum_{k=0}^{n} \frac{1}{k!} \left( 1- \frac{1}{n} \right)\cdots \left( 1- \frac{1-k}{n} \right)\\ \le \sum_{k=0}^{n} \frac{1}{k!} \le \exp(1) $$

Answer 1

Let $N\in\Bbb N$. Then, for $n\geqslant N$,\begin{align}\left(1+\frac1n\right)^n&=\sum_{k=0}^n\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)\\&\geqslant\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right).\end{align}Therefore,$$e\geqslant\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)$$and so$$e\geqslant\lim_{n\to\infty}\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)=\sum_{k=0}^N\frac1{k!}.$$Since this takes place for each $N\in\Bbb N$,$$e\geqslant\sum_{k=0}^\infty\frac1{k!}=\exp(1).$$

Answer 2

This appears as Theorem 8.1.6 in Strichartz's The Way of Analysis. The idea is the following:

• First use the series definition to develop some basic properties of $\exp$: it is strictly positive on $\mathbb{R}$, it is its own derivative, it is strictly increasing, it satisfies $(\exp x)^n = \exp(nx)$.

• Use the inverse function theorem to show that $\exp$ has an inverse function, call it $\ln$, which is differentiable. Use the chain rule to show $\frac{d}{dx} \ln x = \frac{1}{x}$. Also verify that $\ln(1)=0$ and $\ln (x^n) = n \ln x$.

• Write $$ \lim_{n \to \infty} \ln \left(1 + \frac{1}{n}\right)^n = \lim_{n \to \infty} \frac{\ln\left(1 + \frac{1}{n}\right) - \ln(1)}{\frac{1}{n}} = \left.\frac{d}{dx}\right|_{x=1} \ln x = 1.$$

• Take $\exp$ of both sides and use the continuity of $\exp$ to conclude.

Answer 3

So you want to show that $$ \lim_{n\to\infty}\left(1+\frac1n\right)^n=\sum_{k=0}^\infty\frac{1}{k!}=:\exp(1) $$

Let $s_n$ be the partial sum of the series and $t_n=\left(1+\frac1n\right)^n$.

Then the binomial theorem implies that \begin{aligned} 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) \end{aligned}

Thus $t_n\le s_n$, so that $$ \limsup t_n\le \exp(1)\tag{1} $$

On the other hand, if $n\ge m$, then $$ t_{n} \geq 1+1+\frac{1}{2 !}\left(1-\frac{1}{n}\right)+\cdots+\frac{1}{m !}\left(1-\frac{1}{n}\right) \cdots\left(1-\frac{m-1}{n}\right) $$ Let $n\to\infty$, keeping $m$ fixed you get $$ \liminf _{n \rightarrow \infty} t_{n} \geq 1+1+\frac{1}{2 !}+\cdots+\frac{1}{m !}=s_m\tag{2} $$

Combining (1) and (2) you are done.

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Notes. This is Theorem 3.31 in Rudin's Principles of Mathematical Analysis.