definition of the constant $e$

Question

To my knowledge there are two possible ways to define $e^x$

$$e^x = \sum_{i=0}^{\infty}\frac{x^i}{i!}$$

$$e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n$$

So my question is: Why does…

$$\sum_{i=0}^{\infty}\frac{x^i}{i!} = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n$$

Answer 1

The power series $$ \sum_{n\ge0}\frac{x^n}{n!} $$ defines a function on the whole real line, let's call it “exp”. Since power series can be differentiated term by term, we see that $\exp'=\exp$ and also that $\exp0=1$. Consider now the function $$ f(x)=\exp(a-x)\exp x. $$ We have $$ f'(x)=-\exp(a-x)\exp x+\exp(a-x)\exp x=0 $$ so that $f'(x)=0$. Therefore $f$ is constant and its value is $f(0)=\exp a$. So we have $$ \exp(a-b)\exp b=\exp a $$ for all real $a$ and $b$. If $a-b=x$ and $b=y$, we can write the relation as $$ \exp(x+y)=\exp x\exp y $$ and, in particular $\exp(-x)=(\exp x)^{-1}$. So the function $\exp$ never has the value $0$ and so it's always positive. Therefore it's increasing, hence invertible. Denote by $\log$ its inverse. By the inverse function theorem, we have $$ \log'1=1 $$ that is $$ \lim_{h\to0}\frac{\log(1+h)-\log1}{h}=1. $$ However, $\log1=0$ by definition, so, for any $x$, $$ x=\lim_{h\to0}x\frac{\log(1+h)}{h} $$ and, with the substitution $n=x/h$ we get $$ \lim_{n\to\infty}n\log\left(1+\frac{x}{n}\right)=x. $$ For natural $n$, it's clear that $n\log t=\log t^n$ (induction on $\log a+\log b=\log(ab)$ which is the same as $\exp(x+y)=\exp x\exp y$), so we have $$ \lim_{n\to\infty}\log\left(\left(1+\frac{x}{n}\right)^n\right)=x $$ which amounts to say that $$ \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\exp x. $$

Answer 2

HINT : Use the General Binomial Theorem on the expression on the right hand side.

Answer 3

I'm not giving you the full formal proof - you can look that up in just about any calculus textbook - but the basic idea is to use $$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k} \text{ where } \binom{n}{k} = \frac{n!}{k!(n-k)!} $$ to get $$ \left(1 + \frac{x}{n}\right)^n = \sum_{k=0}^n \frac{n!}{n^kk!(n-k)!}x^k = \sum_{k=0}^n \frac{n(n-1)\ldots(n-k+1)}{n^k} \frac{x^k}{k!} \text{.} $$ Then, if $n >> k$ (meaning $n$ is much larger than $k$), you have $$ \frac{n(n-1)\ldots(n-k+1)}{n^k} \approx 1 \text{.} $$ Thus, if you fix $K$, and let $n \to \infty$, the terms involving $x^0,\ldots,x^K$ in $\left(1 + \frac{x}{n}\right)^n$ will converge towards $\frac{x^k}{k!}$.