Radius of convergence of $\sum_{n\ge 1}\frac{z^n}{n}$

Question

In one exercise it is asked to find the radius of convergence of $$\sum\limits_{n\ge 1}\dfrac{z^n}{n};$$ then it is asked to find two values $z_1,z_2\in U=\bigg\{z\in\mathbb{C},|z|=1\bigg\}$ such that $\sum\limits_{n\ge 1}\dfrac{z_1^n}{n}$ diverges and $\sum\limits_{n\ge 1}\dfrac{z_2^n}{n}$ converges.

I've found the Radius $R=1$, $z_1=1$ and $z_2=-1$

Then the correction says the following: it is possible to prove with a more difficult technique that $\sum\limits_{n\ge 1}\dfrac{z^n}{n}$ converges for all $z\in U\backslash \{1\}$. But the correction doesn't provide the result of this proof. Have you got an idea please?

I think this https://en.wikipedia.org/wiki/Dirichlet%27s_test will do the trick. — Nathanael Skrepek, Feb 08 '19 at 11:25
Possible duplicate of Behavior of $\sum_{n=1}^\infty n^{-1}z^n$ on the circle of convergence — Martin R, Feb 08 '19 at 11:33
Show that for $0 < \theta < 2 \pi$ the series $\sum \frac{e^{in \theta}}{n}$ are convergent. — PierreCarre, Feb 08 '19 at 11:43

score 1 · Accepted Answer · answered Feb 08 '19 at 11:25

1

That's a consequence of Dirichlet's test:

$\left(\dfrac1n\right)_{n\in\mathbb N}$ is monotonic;
$\lim_{n\to\infty}\dfrac1n=0$;
$\displaystyle\left(\sum_{n=1}^Nz^n\right)_{N\in\mathbb N}$ is bounded (if $\lvert z\rvert=1$ and $z\neq1$).

answered Feb 08 '19 at 11:25

José Carlos Santos

427,504

Radius of convergence of $\sum_{n\ge 1}\frac{z^n}{n}$

1 Answers1