# 24 Logarithms

## 24.1 The Natural Logarithm

### 24.1.1 Definition: The Natural Logarithm

For a positive number $$x$$ the natural logarithm of $$x$$ is defined as the integral

$\ln x = \int\limits_0^x \frac{1}{t} \ dt$

Additionally, the base of the natural logarithm, denoted $$e$$, is the value such that $$\ln e = 1$$. That is

$1 = \ln e = \int\limits_0^e \frac{1}{t} \ dt$

### 24.1.2 Theorem

#### 24.1.2.1 Lemma 1

$\int\limits_1^{1 + \frac{1}{n}} \frac{1}{1 + \frac{1}{n}}\ dt = \frac{1}{n + 1}$

Proof:

\begin{aligned} \int\limits_1^{1 + \frac{1}{n}} \frac{1}{1 + \frac{1}{n}}\ dt &= \frac{1}{1 + \frac{1}{n}} \cdot t \ \ \Big\rvert_1^{1 + \frac{1}{n}} \\ &= \frac{1}{1 + \frac{1}{n}} \cdot \Big(1 + \frac{1}{n}\Big) - \frac{1}{1 + \frac{1}{n}} \cdot 1 \\ &= 1 - \frac{1}{1 + \frac{1}{n}}\\ &= \frac{1 + \frac{1}{n}}{1 + \frac{1}{n}} - \frac{1}{1 + \frac{1}{n}}\\ &= \frac{\frac{1}{n}}{1 + \frac{1}{n}} \\ &= \frac{\frac{1}{n}}{\frac{n}{n} + \frac{1}{n}} \\ &= \frac{\frac{1}{n}}{\frac{n + 1}{n}} \\ &= \frac{1}{n} \cdot \frac{n}{n + 1} \\ &= \frac{n}{n \cdot (n + 1)} \\ &= \frac{1}{n + 1} \end{aligned}

#### 24.1.2.2 Lemma 2

$\int\limits_1^{1 + \frac{1}{n}} 1\ dt = \frac{1}{n}$

Proof:

\begin{aligned} \int\limits_1^{1 + \frac{1}{n}} 1 \ dt &= t \Big\rvert_{1}^{1 + \frac{1}{n}} \\ &= 1 + \frac{1}{n} - 1 \\ &= \frac{1}{n} \end{aligned}

#### 24.1.2.3 Theorem

$\lim\limits_{n \to \infty} \Big(1 + \frac{1}{n}\Big)^n = e$

Proof:

Let $$t$$ be any real number such that $$\frac{1}{1 + \frac{1}{n}} \leq \frac{1}{t} \leq 1$$. It follows that

\begin{aligned} \int\limits_0^{1 + \frac{1}{n}} \frac{1}{1 + \frac{1}{n}} \ dt &\leq \int\limits_0^{1 + \frac{1}{n}} \frac{1}{t}\ dt &\leq \int\limits_0^{1 + \frac{1}{n}} \frac{1}{1}\ dt \end{aligned}

Using Lemmas 24.1.2.1 and 24.1.2.2, we may write

\begin{aligned} \frac{1}{n + 1} \leq \int\limits_0^{1 + \frac{1}{n}} \frac{1}{t}\ dt \leq \frac{1}{n} \end{aligned}

Next, by applying the definition of the natural logarithm (24.1.1)

\begin{aligned} \frac{1}{n + 1} &\leq \ln \Big(1 + \frac{1}{n}\Big) &\leq \frac{1}{n} \\ \Rightarrow \exp\left\{\frac{1}{n + 1}\right\} &\leq \frac{1}{n + 1} &\leq \exp\left\{\frac{1}{n}\right\} \end{aligned}

Now we isolate the left inequality to show

\begin{aligned} \exp\left\{\frac{1}{n + 1}\right\} &\leq 1 + \frac{1}{n} \\ \Rightarrow \exp\left\{\frac{1}{n + 1}\right\}^{n + 1} &\leq \Big(1 + \frac{1}{n}\Big)^{n + 1} \\ \Rightarrow \exp\left\{\frac{n + 1}{n + 1}\right\} &\leq \Big(1 + \frac{1}{n}\Big)^{n + 1} \\ \Rightarrow e &\leq \Big(1 + \frac{1}{n}\Big)^{n + 1} \end{aligned}

Isolating the right inequality allows us to show

\begin{aligned} 1 + \frac{1}{n} &\leq \exp\left\{\frac{1}{n}\right\} \\ \Rightarrow \Big(1 + \frac{1}{n}\Big)^n &\leq \exp\left\{\frac{1}{n}\right\}^n \\ \Rightarrow \Big(1 + \frac{1}{n}\Big)^n &\leq \exp\left\{\frac{n}{n}\right\} \\ \Rightarrow \Big(1 + \frac{1}{n}\Big)^n &\leq e \end{aligned}

Aligning these two inequalities, we have

\begin{aligned} \Big(1 + \frac{1}{n}\Big)^n &\leq e &\leq \Big(1 + \frac{1}{n}\Big)^{n + 1} \end{aligned}

Notice now that, on the right-most inequality

\begin{aligned} e &\leq \Big(1 + \frac{1}{n}\Big)^{n + 1} \\ \Rightarrow \frac{e}{1 + \frac{1}{n}} &\leq \Big(1 + \frac{1}{n}\Big)^n \end{aligned}

The left-most inequality shows that $$\Big(1 + \frac{1}{n}\Big)^n < e$$, so we may construct the inequality that

\begin{aligned} \frac{e}{1 + \frac{1}{n}} &\leq \Big(1 + \frac{1}{n}\Big)^n &\leq e \end{aligned}

Finally, taking the limit as $$n$$ goes to $$\infty$$

\begin{aligned} \lim_{n \to \infty} \frac{e}{1 + \frac{1}{n}} &\leq \lim_{n \to \infty} \Big(1 + \frac{1}{n}\Big)^n &\leq \lim_{n \to \infty} e \\ \Rightarrow \frac{e}{1 + 0} &\leq \lim_{n \to \infty} \Big(1 + \frac{1}{n}\Big)^n &\leq e \\ \Rightarrow \frac{e}{1} &\leq \lim_{n \to \infty} \Big(1 + \frac{1}{n}\Big)^n &\leq e \\ \Rightarrow e &\leq \lim_{n \to \infty} \Big(1 + \frac{1}{n}\Big)^n &\leq e \end{aligned}

Since $$\lim_{n \to \infty} \Big(1 + \frac{1}{n}\Big)^n$$ is bounded on both sides by $$e$$, $$\lim_{n \to \infty} \Big(1 + \frac{1}{n}\Big)^n = e$$

## 24.2 References

D Joyce, “$$e$$ as the limit of $$(1 + 1/n)^n$$,” Clark University, https://mathcs.clarku.edu/~djoyce/ma122/elimit.pdf