# proof of equivalence of formulas for exp

We present an elementary proof that:

 $\displaystyle\sum_{k=0}^{\infty}\frac{z^{k}}{k!}=\lim_{n\to\infty}\left(1+% \frac{z}{n}\right)^{n}\,.$

There are of course other proofs, but this one has the advantage that it carries verbatim for the matrix exponential and the operator exponential.

At the outset, we observe that $\sum_{k=0}^{\infty}z^{k}/k!$ converges by the ratio test. For definiteness, the notation $e^{z}$ below will refer to exactly this series.

###### Proof.

We expand the right-hand in the straightforward manner:

 $\displaystyle\left(1+\frac{z}{n}\right)^{n}$ $\displaystyle=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{z}{n}\right)^{k}$ $\displaystyle=\sum_{k=0}^{n}\frac{n\cdot(n-1)\cdots(n-k+1)}{n^{k}}\frac{z^{k}}% {k!}=\sum_{k=0}^{n}\pi(k,n)\,\frac{z^{k}}{k!}\,,$

where $\pi(k,n)$ denotes the coefficient

 $\displaystyle 1\left(1-\frac{1}{n}\right)\cdot\left(1-\frac{2}{n}\right)\cdots% \left(1-\frac{k-1}{n}\right)\,.$

Let $\lvert z\rvert\leq M$. Given $\epsilon>0$, there is a $N\in\mathbb{N}$ such that whenever $n\geq N$, then $\sum_{k=n+1}^{\infty}M^{k}/k!<\epsilon/2$, since the sum is the tail of the convergent series $e^{M}$.

Since $\lim_{n\to\infty}\pi(k,n)=1$ for $k$, there is also a $N^{\prime}\in\mathbb{N}$, with $N^{\prime}\geq N$, so that whenever $n\geq N^{\prime}$ and $0\leq k\leq N$, then $\lvert\pi(k,n)-1\rvert<\epsilon/(2e^{M})$. (Note that $k$ is chosen only from a finite set.)

Now, when $n\geq N^{\prime}$, we have

 $\displaystyle\left\lvert\sum_{k=0}^{n}\pi(k,n)\frac{z^{k}}{k!}\,-\,\sum_{k=0}^% {\infty}\frac{z^{k}}{k!}\right\rvert$ $\displaystyle=\left\lvert\sum_{k=0}^{n}(\pi(k,n)-1)\frac{z^{k}}{k!}\,-\,\sum_{% k=n+1}^{\infty}\frac{z^{k}}{k!}\right\rvert$ $\displaystyle\leq\sum_{k=0}^{n}\lvert\pi(k,n)-1\rvert\,\frac{M^{k}}{k!}+\sum_{% k=n+1}^{\infty}\frac{M^{k}}{k!}$ $\displaystyle=\sum_{k=0}^{N}\lvert\pi(k,n)-1\rvert\,\frac{M^{k}}{k!}+\sum_{k=N% +1}^{n}\lvert\pi(k,n)-1\rvert\,\frac{M^{k}}{k!}+\sum_{k=n+1}^{\infty}\frac{M^{% k}}{k!}$ $\displaystyle<\frac{\epsilon}{2e^{M}}\sum_{k=0}^{N}\frac{M^{k}}{k!}+\sum_{k=N+% 1}^{n}\frac{M^{k}}{k!}+\sum_{k=n+1}^{\infty}\frac{M^{k}}{k!}$ (In the middle sum, we use the bound $\lvert\pi(k,n)-1\rvert=1-\pi(k,n)\leq 1$ for all $k$ and $n$.) $\displaystyle<\frac{\epsilon}{2e^{M}}\cdot e^{M}+\frac{\epsilon}{2}=\epsilon\,.\qed$

In fact, we have proved uniform convergence of $\lim_{n\to\infty}\left(1+\frac{z}{n}\right)^{n}$ over $\lvert z\rvert\leq M$. Exploiting this fact we can also show:

 $\displaystyle\left(1+\frac{z}{n}+o\left(\frac{1}{n}\right)\right)^{n}=\left(1+% \frac{z+o(1)}{n}\right)^{n}\to\sum_{k=0}^{\infty}\frac{z^{k}}{k!}\quad\textrm{% (pointwise, as n\to\infty)}$
###### Proof.

$\lvert z\rvert. Given $\epsilon>0$, for large enough $n$, we have

 $\displaystyle\left\lvert\left(1+\frac{w}{n}\right)^{n}-e^{w}\right\rvert<% \epsilon/2\quad\textrm{uniformly for all \lvert w\rvert\leq M.}$

Since $o(1)\to 0$, for large enough $n$ we can set $w=z+o(1)$ above. Since the exponential is continuous11follows from uniform convergence on bounded subsets of either expression for $e^{z}$, for large enough $n$ we also have $\lvert e^{z+o(1)}-e^{z}\rvert<\epsilon/2$. Thus

 $\displaystyle\left\lvert\left(1+\frac{z+o(1)}{n}\right)^{n}-e^{z}\right\rvert% \leq\left\lvert\left(1+\frac{z+o(1)}{n}\right)^{n}-e^{z+o(1)}\right\rvert+% \lvert e^{z+o(1)}-e^{z}\rvert<\epsilon\,.\qed$
Title proof of equivalence of formulas for exp ProofOfEquivalenceOfFormulasForExp 2013-03-22 15:22:52 2013-03-22 15:22:52 stevecheng (10074) stevecheng (10074) 11 stevecheng (10074) Proof msc 30A99 ComplexExponentialFunction ExponentialFunction MatrixExponential