# Compound interest derivation of $e$

I'm reviewing stats and probability, including Poisson processes, and I came across: $$e=\displaystyle \lim_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n$$ I'd like to understand this more fully, but so far I'm struggling. I guess what I'm trying to understand is how you prove that it converges. Can anyone point me toward (or provide) a good explanation of this?

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Sorry for my lack of latex markup skills. The link points to a wikipedia image of this limit properly notated. – ivan Feb 16 '13 at 17:10
There is more than one way to prove this statement. What exactly are you struggling with? – Ludolila Feb 16 '13 at 17:18
This is a common definition of e. What definition are you using? – PyRulez Feb 16 '13 at 17:20

Historically this is the definition of the number $e$. One can show that the sequence

$$\left( 1+\frac{1}{n}\right)^n$$

is increasing and bounded, and thus convergent. We define the limit to be $e$ and then it follows from this limit that $(e^x)'=e^x$.

Here is the proof of it:

Let $a_n= \left( 1+\frac{1}{n}\right)^n$ and $b_n=\left( 1+\frac{1}{n}\right)^{n+1}$. Then clearly $a_n \leq b_n$.

Then $$\frac{a_{n+1}}{a_{n}}=\frac{\left( 1+\frac{1}{n+1}\right)^{n+1}}{\left( 1+\frac{1}{n}\right)^{n}}=\frac{n^n(n+2)^{n+1}}{(n+1)^{2n+1}}=\left( \frac{n(n+2)}{(n+1)^{2}}\right)^{n+1}\frac{n+1}{n}$$

By Bernoulli inequality $$\left(1-\frac{1}{(n+1)^2} \right)^{n+1}\frac{n+1}{n} \geq \left(1-\frac{n+1}{(n+1)^2}\right)\frac{n+1}{n} = 1$$

This shows that $a_n$ is increasing. Similarly $b_n$ is decreasing.

Thus $a_n$ is increasing and bounded by $b_1$, and hence convergent.

Now I will show that its limit is exactly $e$.

Let $l$ be its limit. Then

$$\ln(l) = \lim_n \frac{\ln(1+\frac{1}{n})-\ln 1}{\frac{1}{n}}=\ln'(1)=1$$

Comment: Most textbooks use the last argument to show this limit, but it only works for the wrong reason. We assume that the exponential is differentiable (because it "intuitively" is) and moreover that there is one exponential whose tangent at $(0,1)$ has slope exactly 1 (in other words, we use that $\ln(x)$ is continuous at $x=1$ to define $e$ and $\ln(x)$)..

It is exactly this limit which makes all these work.... And using directly those arguments, the argument really reduces to "because this limit is $e$, it follows that this limit is $e$".

There are actually simpler correct ways of defining $e$ but they all need a deep understanding of Analysis (like viz Taylor Series or using FTC: $\frac{1}{x}$ has an antiderivative).

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Just a quick note on the comment, there is no need to assume that the exponential is differentiable. Simply define the exponential function $\exp(x) = \sum_{i=0}^{\infty}$ and all the usual properties of $\exp(x)$ and $\ln(x)$ follow quite easily, with no need for the FTC. This is particuarly nice because most of the interesting properties of $\exp$ and $\ln$ are differentiable properties rather than integral properties, and they all follow almost immediately from these definitions. – Tom Oldfield Feb 16 '13 at 17:51
@TomOldfield I mentioned that in my last paragraph, but all those definitions use Theorems not covered in an introductory calculus class, at least in North America. ;) And some of them are relatively strong results in Analysis... Moreover, with this approach you define $e=\sum_{n=0}^\infty \frac{1}{n!}=\lim_n \left( 1+\frac{1}{n}\right)^n$, so is not really a new definition, you just prove the existence of the above limit in a different way.... – N. S. Feb 16 '13 at 18:06
@TomOldfield And, to clarify, how would you introduce the exponential this to way to a student if Power series and sequences are NOT part of the material studied in the class? – N. S. Feb 16 '13 at 18:34
Trying to follow the proof, I'm stuck at: $$\frac{\left( 1+\frac{1}{n}\right)^{n}}{\left( 1+\frac{1}{n+1}\right)^{n+1}}=\frac{(n+1)^{2n}}{n^n(n+2)^n}\frac{n+1}{n+2}$$ Can you explain this step? – ivan Feb 16 '13 at 19:28
Oh wait, wait, I think I got it :) – ivan Feb 16 '13 at 19:38

It's not too hard to prove, but it does rely on a few things. (In particular the validity of the taylor expansion of $\ln$ around 1 and that $\exp$ is continuous.)

Consider in general the sequence $n\ln(1+x/n)$ which is defined for all $x$, positive or negative provided $n$ is large enough. (In fact the proof that follows can also be modified slightly to work for complex $x$). Using the power series representation $\ln(1+y) = \sum_{i=1}^\infty (-1)^{i+1}\frac{x^i}{i}$ we get

$$n\ln(1+\frac{x}{n}) = x + \sum_{i=2}^\infty(-1)^{i+1}\frac{x^i}{in^{i-1}}$$

The absolute values of the terms in the series on the right hand side are bounded above by a geometric series (for n large enough) whose summation tends to 0 as $n$ tends to infinity. Hence

$$\lim_{n\rightarrow \infty} n\ln(1+\frac{x}{n}) = x$$

And so using the fact that $\exp(x)$ is a continuous function:

$$\lim_{n\rightarrow \infty}\bigg(1+\frac{x}{n}\bigg)^n = \exp\bigg(\lim_{n\rightarrow \infty}n\ln(1+\frac{x}{n})\bigg) = \exp(x)$$

In particular, set $x = 1$ to get the result you ask for.

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The problem with this type of approach, while it is easy to understand by students, is this: what is your definition of $e$? :) – N. S. Feb 16 '13 at 17:32
BTW: Taylor series are not needed $\lim_{n\rightarrow \infty} n\ln(1+\frac{x}{n}) = x$ follows immediately from the definition of the derivative of $\ln(x)$ at $1$... – N. S. Feb 16 '13 at 17:33
@N.S. We define the function $\exp(x)$ to be the power series $\sum_{n=0}^{\infty} \frac{x^n}{n}$, and $e = \exp(1)$ There is no problem with this definition and allows us to work with both $\exp$ and $\ln$ without the FTC. To derive the limit, it depends on your definition of $\ln(x)$. You can define it as in integral of $\frac{1}{x}$ in which case your method works, or as the inverse of $\exp(x)$ in which case mine does. – Tom Oldfield Feb 16 '13 at 17:40
Yes, you are right there is no problem with this definition, but this is not how most of our undergrad textbooks define $e$. The limit you calculate is actually equivalent to the fact that your definition of the exponential makes sense (i.e. it is an exponential function)... BTW: using the binomial expansion you can prove directly that $\lim_n (1+\frac{1}{n})^n=\sum_{n=0}^\infty \frac{1}{n!}$. – N. S. Feb 16 '13 at 17:45
@N.S. I wouldn't know, this is how I had $e$ defined to me as an undergraduate. Personally I would view the real test of a function as being exponential as having the property $f(x+y)=f(X)f(y)$. Yes, this definition can be derived from the other, and that just goes to show that it doesn't really matter which one you use, as long as you can use it easily! I guess everyone should just pick their favourite :) – Tom Oldfield Feb 16 '13 at 18:03

I have removed the more precise, but more confusing, justifications of $(1)$ and $(2)$. In this answer, I show that $$\left(1+\frac1{n+1}\right)^{n+1}\ge\left(1+\frac1n\right)^n\tag{1}$$ and $$\left(1+\frac1{n+1}\right)^{n+2}\le\left(1+\frac1n\right)^{n+1}\tag{2}$$ Therefore, $(1)$ says that $\left(1+\frac1n\right)^n$ is an increasing sequence and $(2)$ says that $\left(1+\frac1n\right)^{n+1}$ is a decreasing sequence. Since $\left(1+\frac1n\right)^n\le\left(1+\frac1n\right)^{n+1}$ the increasing sequence is bounded above and the decreasing sequence is bounded below; thus, they both converge.

Since $\lim\limits_{n\to\infty}\frac{\left(1+\frac1n\right)^{n+1}}{\left(1+\frac1n\right)^n}=\lim\limits_{n\to\infty}\left(1+\frac1n\right)=1$. They both converge to the same limit. This is called $e$. $$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{3}$$

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A great elementary proof is on Rudin's Principles of Mathematical Analysis page 64 Theorem 3.31.

PS: Sorry I don't know how to stretch it, maybe you can download the pic and enlarge.

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Please write it down here for us who don't have that book at hand. – vonbrand Feb 16 '13 at 18:59

I want to put this as an answer rather than a comment so that people can find it easily:

https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function#Equivalence_of_characterizations_1_and_2

^ Wikipedia has the cleanest approach I can find. i.e. It doesn't pull anything out of the blue, and it doesn't sit on top of other advanced results.

I'm not going to replicate the working; I am confident it will not change for the worse.

EDIT: I just noticed it cites its source as Rudin, theorem 3.31, p. 63–5, so it is the same as the picture posted in another answer!

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