proof of equivalence of formulas for exp

We present an elementary proof that:

${\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{z^k}{k!}}=\underset{n\to \mathrm{\infty }}{lim}{\left(1+{\displaystyle \frac{z}{n}}\right)}^n.$

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${\sum }_{k=0}^{\mathrm{\infty }}{z}^k/k!$ 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:

${\left(1+{\displaystyle \frac{z}{n}}\right)}^n$$={\displaystyle \sum _{k=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left({\displaystyle \frac{z}{n}}\right)}^k$

$={\displaystyle \sum _{k=0}^n}{\displaystyle \frac{n\cdot (n-1)\mathrm{\cdots }(n-k+1)}{n^k}}{\displaystyle \frac{z^k}{k!}}={\displaystyle \sum _{k=0}^n}\pi (k,n){\displaystyle \frac{z^k}{k!}},$

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

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

Let $|z|\le M$.Given $ϵ>0$, there is a $N\in \mathbb{N}$ such that whenever $n\ge N$, then,since the sum is the tail of the convergent series $e^M$.

Since ${lim}_{n\to \mathrm{\infty }}\pi (k,n)=1$ for $k$, there is also a $N^{\prime }\in \mathbb{N}$, with $N^{\prime }\ge N$, so that whenever $n\ge N^{\prime }$ and $0\le k\le N$, then.(Note that $k$ is chosen only from a finite set.)

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

$\left|{\displaystyle \sum _{k=0}^n}\pi (k,n){\displaystyle \frac{z^k}{k!}}-{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{z^k}{k!}}\right|$$=\left|{\displaystyle \sum _{k=0}^n}(\pi (k,n)-1){\displaystyle \frac{z^k}{k!}}-{\displaystyle \sum _{k=n+1}^{\mathrm{\infty }}}{\displaystyle \frac{z^k}{k!}}\right|$

$\le {\displaystyle \sum _{k=0}^n}|\pi (k,n)-1|{\displaystyle \frac{M^k}{k!}}+{\displaystyle \sum _{k=n+1}^{\mathrm{\infty }}}{\displaystyle \frac{M^k}{k!}}$

$={\displaystyle \sum _{k=0}^N}|\pi (k,n)-1|{\displaystyle \frac{M^k}{k!}}+{\displaystyle \sum _{k=N+1}^n}|\pi (k,n)-1|{\displaystyle \frac{M^k}{k!}}+{\displaystyle \sum _{k=n+1}^{\mathrm{\infty }}}{\displaystyle \frac{M^k}{k!}}$

(In the middle sum, we use the bound $|\pi (k,n)-1|=1-\pi (k,n)\le 1$ for all $k$ and $n$.)

In fact, we have proved uniform convergence of ${lim}_{n\to \mathrm{\infty }}{\left(1+\frac{z}{n}\right)}^n$over $|z|\le M$.Exploiting this fact we can also show:

${\left(1+{\displaystyle \frac{z}{n}}+o\left({\displaystyle \frac{1}{n}}\right)\right)}^n={\left(1+{\displaystyle \frac{z+o(1)}{n}}\right)}^n\to {\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{z^k}{k!}}\mathit{\hspace{1em}}\text{(pointwise, as }n\to \mathrm{\infty }\text{)}$

Proof.

.Given $ϵ>0$, for large enough $n$, we have

Since $o(1)\to 0$, for large enough $n$ we can set $w=z+o(1)$ above.Since the exponential is continuous, for large enough $n$ we also have . Thus