I'm hoping someone can help me with the following question from Stein and Shakarchi Volume 1 on Fourier Analysis (p. 24):

For $z \in \mathbb C$, we define the complex exponential by $$\sum_{n=0}^{\infty} \frac{z^n}{n!}.$$ Prove that the above definition makes sense, by showing the series converges for every complex number $z$. Moreover, show that the convergence is uniform on every bounded subset of $\mathbb C$.

I know that convergence is easily shown with the ratio test or an absolute convergence argument, but I'm trying to prove convergence formally, and keep getting stuck. So far, I've tried the following. A series converges if the sequence of partial sums $$\sum_{j=0}^{n} \frac{z^j}{j!}$$ converges as $n \to \infty$. So showing that $$\lim_{n \to \infty} \left| \sum_{j=0}^{n} \frac{z^j}{j!} - e^z \right| = 0$$ appears to be the goal. We could also use the $N-\epsilon$ definition, which may help, or show the sequence is Cauchy. But I can't seem to get any of these to work. With the uniform convergence, I seem to run into the same problems.

Any help or hints would be appreciated.

  • $\begingroup$ Note that $|e^z| \le e^{|z|}$, so the problem of convergence reduces to the real case. $\endgroup$
    – copper.hat
    Nov 16, 2016 at 3:42

1 Answer 1



Note that for any fixed $z$, we have for $M>N>|z|$

$$\begin{align} \left|\sum_{n=N}^M \frac{z^n}{n!}\right|&\le \sum_{n=N}^M \frac{|z|^n}{n!}\\\\ &\le \sum_{n=N}^M \frac{|z|^n}{N^n}\\\\ &=\frac{|z/N|^N -|z/N|^{M+1}}{1-|z/N|} \end{align}$$

  • $\begingroup$ By writing the sum to $\infty$, you are assuming that it converges. You need to make the upper limit $m$ and show that the sum converges as $m \to \infty$. $\endgroup$ Nov 16, 2016 at 5:09
  • $\begingroup$ @martycohen Actually, the upper limit implies that a limit is taken. In fact, the last step actually evaluates that implied limit. Does that make sense? $\endgroup$
    – Mark Viola
    Nov 16, 2016 at 5:15
  • $\begingroup$ I would make the upper limit $m$, where $m > N$. The numerator of the fraction would then be $|z/N|^N-|z/N|^{m+1} = |z/N|^N(1-|z/N|^{m-N+1})$ and show that that expression goes to zero. $\endgroup$ Nov 16, 2016 at 5:22
  • $\begingroup$ @martycohen Marty, I've edited to incorporate your suggestion. Much appreciative. -Mark $\endgroup$
    – Mark Viola
    Nov 17, 2016 at 4:50
  • $\begingroup$ Thanks a lot. I assume this goes to showing uniform convergence right? I just don't really see how... I may be missing a theorem somewhere... $\endgroup$
    – K.Reeves
    Nov 18, 2016 at 13:20

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