# Show that $e^{x+y}=e^xe^y$ using $e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$.

I was looking for a proof of $e^{x+y}=e^xe^y$ using the fact that $$e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n.$$

So I have that $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{(x+y)^k}{n^k}=\sum_{k=0}^n\frac{1}{n^k}\sum_{i=0}^k\binom{k}{i}x^iy^{k-i}=\sum_{k=0}^n\binom{n}{k}\sum_{i=0}^k\binom{k}{i}\left(\frac{x}{n}\right)^i\left(\frac{y}{n}\right)^{k-i}$$

But I can't get $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{x}{n}\right)^k\sum_{i=0}^n\binom{n}{i}\left(\frac{y}{n}\right)^i.$$

Any idea ?

• try the other way $\lim_{n \to \infty} (1+\frac{x}{n})^n(1+\frac{y}{n})^n$. Then ignore the higher order terms – user265328 Sep 25 '15 at 16:50
• I got it, but I would like to conclude like this if it's possible :-) – idm Sep 25 '15 at 16:54
• I think it does not matter. Anyways I will try to explain it to you without writing it mathematically. While going in the direction you want, we have to add and substract the terms we neglected to make $\lim_{n\to \infty} (1 + \frac{x}{n})^n(1+\frac{y}{n})^n = \lim_{n\to \infty} (1+\frac{x+y}{n})^n$. Then we will be left with two terms $(1+\frac{x+y}{n})^n - (something)$. This something has powers of $n$ in the denominator and will go to $0$ after taking the limit. I personally do not like this approach. – user265328 Sep 25 '15 at 17:11

My favorite approach to these problems is to first prove:

Lemma: If $g$ is a real-valued function on the natural numbers such that $$\lim_{n\to\infty}n(g(n)-1)=c$$ then: $$\lim_{n\to\infty} g(n)^{n} = e^c$$

Proof:

Case c=0:

Restrict to $n$ so that $n|g(n)-1|<1$. Then: \begin{align} g(n)^n &= (1+(g(n)-1))^n\\ & = 1+ \sum_{k=1}^{n}\binom{n}{k}(g(n)-1)^k\end{align} Since $\binom{n}{k}\leq \frac{n^k}{k!}$ and $\frac{1}{k!}\leq \frac{1}{2^{k-1}}$:

\begin{align} |g(n)^n-1| &\leq \sum_{k=1}^{n}\frac{1}{k!}\left(n|g(n)-1|\right)^k\\ &\leq n|g(n)-1|\sum_{k=1}^\infty \frac{1}{2^{k-1}}\\ &=2n|g(n)-1| \end{align}

So $\left|g(n)^n-1\right|\to 0$.

General case:

Let $h(n)=\frac{g(n)}{1+c/n}$. Then $$n(h(n)-1)=\frac{n(g(n)-1)-c}{1+c/n}\to 0.$$ So, by the case $c=0$, we have that $h(n)^n\to 1$. But we know that $(1+c/n)^n\to e^c$, so we are done.

The case $c=0$ used only binomial theorem and basic inequalities. The general case requires knowledge that $\lim_{n\to\infty}\left(1+\frac{c}{n}\right)^n\to e^c.$

The lemma can be rewritten in little-$o$ notation as:

If $g(n)=1+\frac{c}{n}+o\left(\frac1n\right)$ then $g(n)^n\to e^c$.

The problem at hand

Next, show that, for fixed $x,y$ that: \begin{align}G(n)&=\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right) \end{align}

satisfies the property that $n(G(n)-1)\to x+y$ since $$n(G(n)-1)=x+y+\frac{xy}{n}.$$

This shows what you wanted.

Another application: Euler's formula

If $h(x)$ is a function such that $h(x)=1+cx+o(x)$ then $h(x/n)^n\to e^{cx}$. This condition $h(x)=1+cx+o(x)$ is equivalent to $h(0)=1,h'(0)=c$.

One interesting case is $h(x)=\cos x+i\sin x$. We can show purely geometrically that $\cos x+i\sin x= 1+ix+o(x)$, and by induction prove that $h(x)^n=h(nx)$, so we have that $$\cos x+i\sin x =h(x)= \lim_{n\to\infty} h(x/n)^n=\lim_{n\to\infty}\left(1+\frac{ix}n\right)^n$$

The geometric proof is roughly, for all $x\in(0,\pi/2)$:

1. $\sin x<x$. This is because $\frac{1}{2}\sin x$ is the area of a triangle contained in the circle wedge of angle $x$ which as area $x/2$.

2. $0\leq 1-\cos x<\frac{1}{2}x^2$. This is true because $1-\cos x = 2\sin^2(x/2)<\frac{x^2}{2}$ by step (1).

3. $x<\tan x$. This is because the wedge of angle $x$ is contained in the triangle $(0,0)$, $(1,0)$, $(1,\tan x)$ of area $\frac{1}{2}\tan x$.

4. From (3) and (1), we get $x\cos x< \sin x<x$. But $\cos x =1+o(x)$ so $x\cos (x)=x+o(x^2)$ and thus $\sin(x)=x+o(x^2)$.

This shows that $\cos x +i\sin x = 1+ix + o(x)$.

• what is $g(n)$ ? – idm Sep 25 '15 at 17:03
• Any function with the property that I've stated. Edited the answer to make that clear. – Thomas Andrews Sep 25 '15 at 17:07
• Well done! ... +1 – Mark Viola Sep 25 '15 at 21:01
• Beautiful proof! – Hasan Saad Sep 25 '15 at 23:47
• In past I had shown that $(1 + x/n)^{n}(1 + y/n)^{n} - (1 + (x + y)/n)^{n}$ tends to $0$ (see math.stackexchange.com/a/541330/72031), but it appears that showing $a/b \to 1$ is easier here than showing $a - b \to 0$. Plus derivation of the final result involving $\cos x + i\sin x$ is truly a gem. +1 – Paramanand Singh Sep 27 '15 at 10:37

\begin{align*} e^x e^y &= \lim_{n\rightarrow \infty}(1+\frac{x}{n})^n\lim_{n\rightarrow \infty}(1+\frac{y}{n})^n\\ &=\lim_{n\rightarrow \infty}(1+\frac{x}{n})^n(1+\frac{y}{n})^n\\ &=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n}+\frac{xy}{n^2})^n\\ &=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n \frac{(1+\frac{x+y}{n}+\frac{xy}{n^2})^n}{(1+\frac{x+y}{n})^n}\\ &=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n \left(1+\frac{xy}{n^2(1+\frac{x+y}{n})}\right)^n\\ &=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n\left(1+\frac{xy}{n^2(1+\frac{x+y}{n})}\right)^{\frac{n^2(1+\frac{x+y}{n})}{xy}\frac{xy}{n^2(1+\frac{x+y}{n})}n}\\ &= \lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n\\ &=e^{x+y}. \end{align*}

• It would facilitate to use $$1+\frac1n\le \left(1+\frac1{n^2}\right)^n\le 1+\frac2n$$and squeeze the second term in Lines $3-5$. – Mark Viola Sep 26 '15 at 15:48