Definition – Euler’s constant

Contents
1. Definitions
2. $e_n$ is an increasing sequence
2. $e_n$ converges
3. The derivative of $e^x$

1. Definition

Euler’s constant, written $e$ , is defined as the limit,

$e = \lim_{n\rightarrow \infty} (1+\frac{1}{n} )^n = 2.7182818...$

2. $e_n$ is an increasing sequence

Recall that, for a set of $n$ real numbers, their geometric mean is less than or equal to the arithmetic mean. Now consider the set of $(n+1)$ numbers,

$\underbrace{(1+\frac{1}{n}), ..., (1+\frac{1}{n})}_{\text{n times}}, 1$

Now, based on their geometric vs arithmetic means,

$\sqrt[n+1]{(1+\frac{1}{n})^n} \leq \frac{n+2}{n+1} = 1 + \frac{1}{n+1}$

Finally, raising both sides to the power of $n+1$ , we obtain,

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

Ie, $e_n \leq e_{n+1}$ . $\square$

3. $e_n$ converges

It can be proven that $e$ has an upper bound (pick any $n\in\mathbb{N}$ such that $n \geq 3$ ). Since $e_n$ is increasing and bounded above, it follows that $e_n$ must converge to some limit.

3. The derivative of $e^x$

Theorem – $\frac{d}{dx} e^x = e^x$

Proof –

We first use the definition of the derivative of a function $f(x) = e^x$ . From this definition, we have,

$\begin{array}{ccc} f'(x) &= &\lim_{h\rightarrow 0} \frac{e^{x+h} - e^x}{h} \\\\ &= &e^x \lim_{h\rightarrow 0} \frac{e^h - 1}{h} \\\\ &= & f(x) \lim_{h\rightarrow 0} \frac{e^h - 1}{h} \end{array}$

It now remains to determine the value of $\lim_{h\rightarrow 0} \frac{e^h - 1}{h}$ . We have two cases. The first case is if $h = \frac{1}{n}$ for some $n\in\mathbb{N}$ and the second is when $h$ cannot be written in this form, but is still an element of the real numbers.

To prove the first case, $h = \frac{1}{n}$ . Then, note that

$(1+\frac{1}{n+1})^n \leq e \leq (1+\frac{1}{n})^{n+1} \leq (1+\frac{1}{n-1})^n$

For all $n\in\mathbb{N}$ . Ignoring the 3rd expression, and taking the $n$ -th root of all three remaining expressions,

$1+\frac{1}{n+1} \leq e^{1/n} \leq 1+\frac{1}{n-1} \iff \frac{1}{n+1} \leq e^{1/n} -1 \leq \frac{1}{n-1}$

But, by the sandwich theorem for limits, we have that $\lim_{n\rightarrow \infty} e^{1/n} -1 = 0$ , so finally by the algebra of limits, $\lim_{n\rightarrow \infty} e^{1/n} =1$ . This proves the first case.

Secondly, if $h\in\mathbb{R}^+$ cannot be written in the form $\frac{1}{n}$ , then by the Archimedes principle, there exists an integer $k$ such that

$\frac{1}{n-1} \leq h \leq \frac{1}{n} \implies h(n-1) \leq 1 \leq hn \quad \text{(1)}$

Then, taking the reciprocal of each terms, we have $n-1 \leq \frac{1}{h} \leq n$ , and so $latex $. Then, recall that

$(1+\frac{1}{n})^n \leq e \leq (1+ \frac{1}{n-2})^{n-1} \implies (1+\frac{1}{n})^{hn} \leq e^h \leq (1+\frac{1}{n-2})^{h(n-1)} \quad \text{(2)}$

Due to $(1)$ and $(2)$ , we have

$1+\frac{1}{n} \leq e^h \leq 1+\frac{1}{n-2} \implies \frac{1}{n} \leq e^h -1 \leq \frac{1}{n-2}$

And again, by the sandwich theorem, we have $\lim_{h\rightarrow 0} e^h = 1$ . Combining the result for both cases, we can finally confirm that,

$f'(x) = f(x) \lim_{h\rightarrow 0} \frac{e^h - 1}{h} = f(x) = e^x\quad\square$