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