we initially start with one dollar in the bank account, that is, $P=1$.
the interest rate is $100\%$, that is, the bank will double our money at every compound period. This would mean $r=1$.
we are interested in our bank balance after one year, that is, $t=1$.

For this simple case, our formula for compound interest becomes:

$$\begin{align*} A=\Big(1+\frac{1}{n}\Big)^{n} \end{align*}$$

The only variable here is $n$, which is the number of compounds per year. For instance, $n=1$ for annual compounds, whereas $n=4$ for quarterly compounds. Let's now compute $A$ for different values of $n$ like so:

$n$	$A$	$n$	$A$
1 (annually)	2.0000	52 (weekly)	2.6926
2 (bi-annually)	2.2500	365 (daily)	2.7146
4 (quarterly)	2.4414	8760 (hourly)	2.7181
12 (monthly)	2.6130	525600 (per min)	2.7183

Note that the values of $A$ are all rounded to four decimal places. We can see that as $n$ increases, the value of $A$ also increases, but the amount of increase slows down. For instance, increasing $n$ from $1$ to $2$ gives us an additional $\$0.25$, but increasing $n$ from $365$ to $8760$ only gives us around $\$0.003$ additionally.

As a result, $A$ seems to be converging to some constant of around $2.718$; this constant is the famous Euler's constant $e$. The larger the $n$, the closer we approach $e$, hence it makes sense to define $e$ as the limit of $A$ as $n$ tends to infinity.

Example.

Formal definition of Euler's constant

Euler's constant $e$ is defined as follows:

$$e=\lim_{n\to\infty} \Big[\Big(1+\frac{1}{n}\Big)^n\Big]$$

Note that $e$ is irrational:

$$e=2.71828...$$

Let's now expand on the formal definition of Euler's constant and express $e^\lambda$ as a limit where $\lambda$ is some scalar number.

Theorem.

Expressing the power of Euler's constant as limits

If $\lambda\in\mathbb{R}$, then:

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

Proof. From the basic rules of taking powers, we have that:

$$\Big(1+\frac{\lambda}{n}\Big)^{n} =\Big[\Big(1+\frac{\lambda}{n}\Big)^{n/\lambda}\Big]^\lambda $$

Now, let's take the limit as $n$ tends to infinity:

$$\begin{equation}\label{eq:yTGF12ryu589R5LxGty} \lim_{n\to\infty}\Big(1+\frac{\lambda}{n}\Big)^{n} = \lim_{n\to\infty}\Big[\Big(\Big(1+\frac{\lambda}{n}\Big) ^{n/\lambda}\Big)^\lambda\Big] \end{equation}$$

We now want to show that the right-hand side is equal to $e^\lambda$. Let's define a new variable:

$$k=\frac{n}{\lambda}$$

Notice how $k$ also tends to infinity as $n$ tends to infinity. Rewriting the right-hand side of \eqref{eq:yTGF12ryu589R5LxGty} in terms of $k$ gives:

$$\lim_{k\to\infty}\Big[\Big(\Big(1+\frac{1}{k}\Big) ^{k}\Big)^\lambda\Big]$$

Using the power rule of limits, we can take the power of $\lambda$ outside the limit:

$$\begin{equation}\label{eq:PxSqujAPFRRw0WW1Wlc} \begin{aligned}[b] \lim_{k\to\infty} \Big[\Big(\Big(1+\frac{1}{k}\Big)^{k}\Big)^\lambda\Big] &=\Big[ \lim_{k\to\infty} \Big(\Big(1+\frac{1}{k}\Big)^{k}\Big)\Big]^\lambda\\ &=e^\lambda\\ \end{aligned} \end{equation}$$

Substituting \eqref{eq:PxSqujAPFRRw0WW1Wlc} into \eqref{eq:yTGF12ryu589R5LxGty} gives us the desired identity:

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

This completes the proof.

■

Published by Isshin Inada

Edited by 0 others

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