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.
Answer
You can use following steps.
- 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 $m = \max (\vert a \vert, \vert b \vert)$
- For $u \in \mathbb C$ you have $$\left\vert e^u-(1+u) \right\vert \le \sum_{k=2}^{+\infty} \frac{\vert u \vert^k}{k!} \le \vert u \vert^2 \sum_{k=0}^{+\infty} \frac{\vert u \vert^k}{k!}=\vert u \vert^2 e^{\vert u \vert} \tag{2}$$
- 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 $$\left\vert e^{ku} -(1+u)^k\right\vert \leq\frac{\vert k u \vert^2 e^{\vert ku \vert}}{k} \tag{3}$$
- Finally for $z \in \mathbb{C}$ and denoting $u=\frac{z}{n}$ and $k=n$, we obtain using (3) $$\left\vert e^z -\left(1+\frac{z}{n}\right)^n \right\vert \le \frac{\vert z \vert^2 e^{\vert z \vert}}{n} \tag{4}$$
- 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 $\mathbb C$.
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