prove that $\sum_{n=1}^{\infty} \frac{z^n}{n!}$ does not converge uniformly to $e^z$ on $\mathbb{C}$

Question

How can I prove that $\sum_{n=1}^{\infty} \frac{z^n}{n!}$ does not converge uniformly to $e^z$ on $\mathbb{C}$?

My intuition is to assume, for a contraction, that a power series converges uniformly on $\mathbb{C}$. Then, let $$S_{m}(z) = \left(1+\frac{z}{1!}+\frac{z^{2}}{2!}+···+\frac{z^{m}}{m!}\right)−e^{z}.$$ Consider $$|S_{K+1}(z) − S_{K}(z)|$$ as $z \to \infty$ to get a contradiction.

Can I get some help in formulating the details or any other more elegant approach in this?

Answer 1

Set $S_{n}(z)=\displaystyle\sum_{k=1}^{m}\dfrac{z^{k}}{k!}$, if it were uniformly convergent, then for some $N$, for all $z\in{\bf{C}}$, $|S_{m}(z)-S_{n}(z)|<1$ for all $m>n\geq N$, by setting $m=N+1$ and $n=N$, then $\left|\displaystyle\dfrac{z^{N+1}}{(N+1)!}\right|<1$ for any $z\in{\bf{C}}$. Realizing $z=x>0$, then $0<\dfrac{x^{N+1}}{(N+1)!}<1$, and taking $x\rightarrow\infty$ then the expression blows up.

Answer 2

If uniformly convergent then for any $\epsilon > 0$, we have for all $m > n$ sufficiently large and all $z$

$$\left|\sum_{k=n+1}^m \frac{z^k}{k!}\right| < \epsilon.$$

Take $m = 2n$ and $z = (2n)^2$ and we have

$$\left|\sum_{k=n+1}^m \frac{z^k}{k!}\right| = \sum_{k=n+1}^{2n} \frac{(2n)^{2k}}{k!} \geqslant n \cdot \frac{(2n)^{2n}}{(2n)!} \to_{n \to \infty} \infty$$

Answer 3

If $S_m(z)\rightarrow 0$ uniformly on $\mathbb C,$ then there is an $n$ such that $|S_n(z)|<1, z \in \mathbb C.$ Since a bounded entire function is constant, $e^z$ is a polynomial, which we know to be false.