Find the region in the complex plane $z$ where the series $e^{-nx}$ converges uniformly; explain

Question

Find the region in the complex plane z where the series $\sum_{n=0}^\infty e^{-nz} $ converges uniformly; explain.

I know I need to perform the ratio test, but from there, I'm not sure how to determine where the series converges uniformly.

Answer 1

As a suggestion, try realizing the sum as a geometric series: $$ e^{-nz} = \big(e^{-z}\big)^n. $$ The series can then be seen to converge absolutely wherever $|e^{-z}| < 1$, or equivalently where $|e^z|= e^{\mathrm{Re} (z)} > 1$. Hence we see that the series converges uniformly on every set of the form $\{z:\mathrm{Re}(z) \ge \epsilon>0\}$ by the Weierstrass $M$-test.

To get a more precise handle on where the series converges uniformly, the $N$th partial sum of the series then has an explicit formula given by $$ s_N(z) = \sum_{n=0}^N\big(e^{-z}\big)^n = \frac{1-\big(e^{-z}\big)^{N+1}}{1-e^{-z}}, $$ so the question is to find where the sequence of functions $s_N(z)$ converges uniformly, which is where the sequence $\{s_N(z)\}$ is uniformly Cauchy, so try playing with estimates applied to the difference $|s_{N}(z)-s_{N+p}(z)|$.

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Though we get uniform convergence on each closed half-plane $\{z:\mathrm{Re}(z)\ge \epsilon > 0\}$, we do not get that the series converges uniformly on the open half-plane $\{z:\mathrm{Re}(z) > 0\}$. For if we did, then we could execute the following exchange of limits (cf. Rudin's Principles of Mathematical Analysis, Theorem 7.11): $$ \lim_{N\to\infty}\lim_{t\to 0}s_N(t) =\lim_{t\to 0}\lim_{N\to\infty}s_N(t). $$ In particular, $\lim_{t\to 0}s_N(t)$ converges as $N\to\infty$, but $\lim_{t\to 0}s_N(t)=s_N(0) = N+1$ clearly blows up as $N\to\infty$.