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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$\begingroup$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).
$\endgroup$ 16 $\begingroup$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.
$\endgroup$ 7 $\begingroup$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 $e$. $$ e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{3} $$
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$I want to put this as an answer rather than a comment so that people can find it easily:
^ 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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