Do the polynomials $(1+z/n)^n$ converge compactly to $e^z$ on $\mathbb{C}$?

The question is

Do the polynomials $p_n(x)=(1+z/n)^n$ converge compactly (or uniformly on compact subsets) to $e^z$ on $\mathbb{C}$?

I thought about expanding $$p_n(z)=\sum_{k=0}^n a_k^{(n)}z^k$$ where $$a_k^{(n)}=\binom{n}{k}\frac{1}{n^k}=\frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)$$ and trying to show that $\frac{1}{k!}-a_k^{(n)}$ decreases sufficiently fast on any closed ball. That is, I tried to show $$\lim_{n\rightarrow\infty}\max_{z\in\overline{B_0(A)}}\left|\sum_{k=0}^n\frac{z^k}{k!}-p_n(z)\right|=0$$ for any fixed $A>0$, but I had difficulty with this approach.

Any help is appreciated.

Solution 1:

You can use following steps.

1. For $a, b \in \mathbb C$ and $k \in \mathbb N$ you have $$\vert a^k -b^k \vert =\vert a-b \vert \vert a^{k-1}+b a^{k-2}+\dots+b^{k-1}\vert\le \vert a - b \vert k m^{k-1} \tag{1}$$ where $m = \max (\vert a \vert, \vert b \vert)$
2. For $u \in \mathbb C$ you have $$\left\vert e^u-(1+u) \right\vert \le \sum_{k=2}^{+\infty} \frac{\vert u \vert^k}{k!} \le \vert u \vert^2 \sum_{k=0}^{+\infty} \frac{\vert u \vert^k}{k!}=\vert u \vert^2 e^{\vert u \vert} \tag{2}$$
3. Now taking $a=e^u,b=1+u$, we get $m=\max(\vert e^u \vert,\vert 1+u \vert) \le \max(e^{\vert u \vert},1+\vert u \vert) \le e^{\vert u \vert}$. For $k \ge 1$ applying (1) and (2) successively, we get $$\left\vert e^{ku} -(1+u)^k\right\vert \leq\frac{\vert k u \vert^2 e^{\vert ku \vert}}{k} \tag{3}$$
4. Finally for $z \in \mathbb{C}$ and denoting $u=\frac{z}{n}$ and $k=n$, we obtain using (3) $$\left\vert e^z -\left(1+\frac{z}{n}\right)^n \right\vert \le \frac{\vert z \vert^2 e^{\vert z \vert}}{n} \tag{4}$$
5. For $K \subset \mathbb C$ compact, one can find $M > 0$ such that $M \ge \sup\limits_{z \in K} \vert z \vert$ which implies $$\sup\limits_{ z \in K} \left\vert e^z -\left(1+\frac{z}{n}\right)^n \right\vert \le \frac{M^2 e^{M}}{n} \tag{5}$$ proving that $(p_n)$ converges uniformly to $e^z$ on every compact subset of $\mathbb C$.

Solution 2:

HINT: It's not that hard from a certain point of view:

Take $\epsilon > 0$. Consider $N$ natural so that $\sum_{k > N} \frac{A^k}{k!} < \epsilon/3$. Let now $n_{\epsilon}$ so that $$\sum_{k=0}^N\frac{1}{k!}\left |\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)-1\right|A^k< \epsilon/3 $$ for $n > n_{\epsilon}$.

Now, for $n > n_{\epsilon}$ we have the estimate

\begin{eqnarray} |e^z - p_n(z)|\le \sum_{k=0}^N\frac{1}{k!}\left |\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)-1\right|A^k + 2 \sum_{k > N} \frac{A^k}{k!}< \epsilon \end{eqnarray} for all $z$ so that $|z| \le A$.

${\bf Added:}$ Similarly one can prove the following:

Given a sequence of series $( \sum_{k \ge 0} \phi_k^{(n)} (z))_n $ so that $\phi_k^{(n)}\to \phi_k$ uniformly on compacts and on every compact the series $( \sum_{k \ge 0} |\phi_k^{(n)} (z)|))_n$ are all termwise bounded by a convergent series, then the sums of the series $( \sum_{k \ge 0} \phi_k^{(n)} (z))_n $ converge uniformly on compacts to $\sum_{k \ge 0} \phi_k (z)$.

This is more or less a dominated convergence theorem.

Solution 3:

Here's a less computational way to get the result.

Lemma 1: If $f_n: E \to \mathbb C $ are uniformly bounded and converge uniformly to $f$ on $E,$ then $e^{f_n} \to e^f$ uniformly on $E.$ Proof: Easy.

Lemma 2: Let $\log $ denote the principal value logarithm. Then there is a constant $C$ such that for $0<|u|< 1/2,$

$$\left|\frac{\log (1+u)}{u}-1\right| \le C|u|.$$

Proof: Inside the absolute values on the left we have a function that extends to be analytic in $D(0,1)$ with value $0$ at $0.$ The result follows easily.

So let $R>0.$ For $0<|z|<R$ and $n>2R,$ Lemma 2 shows

$$|n\log(1+z/n) - z| = |z|\,\left|\frac{\log(1+z/n)}{z/n}-1\right| \le R\cdot C(R/n) = CR^2/n.$$

So we have uniform convergence of $n\log(1+z/n)$ to $z$ on $D(0,R),$ at least along the sequence of $n$'s greater than $2R.$ Because these functions are uniformly bounded on $D(0,R),$ Lemma 1 shows

$$\tag 1 \exp (n\log(1+z/n)) \to e^z$$

uniformly on $D(0,R).$ Since the left side of $(1)$ equals $[\exp (\log(1+z/n))]^n = p_n(z),$ we're done.