Determine the radius of convergence 
of the power series:
![\[ \sum_{n=0}^{\infty} \frac{z^n}{(n+1)2^n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6cfaad8a430bb8ec14bc94c24f8fb366_l3.png "Rendered by QuickLaTeX.com")
Test for convergence at the boundary points if is finite.
Let 
. Then we have
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{z^{n+1}}{(n+2)2^{n+1}} \right) \left( \frac{(n+1)2^n}{z^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{(n+1)z}{2(n+2)} \\ &= \frac{z}{2}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-108a4bc397b8b971bd060f9d27b30872_l3.png "Rendered by QuickLaTeX.com")
This implies 
. So, the series converges for all 
. Furthermore, the series is convergent at every boundary point 
except 
by Dirichlet’s test (Theorem 10.17). Hence, the series converges for all 
with 
.
Minor typo: -pi < argz_1 + argz_2 <= pi
Minor placement of comment error: see below
I don’t recall any examples using |z| other then 1 so the boundary points |z|=2, z≠2 satisfied using Dirichlet’s test is new to me. I assume using z^n = r^n(cos nθ + i sin nθ) with θ=0 is the key, separating the series into real and imaginary parts of which the real part diverges. But at θ=π would we not be using Leibniz’s instead of Dirichlet’s test to prove convergence?
Some more thoughts: Apostol theorem 10.19 stipulates θ not equal to multiples of π, so Dirichlet’s test can’t be used at z=-1? So then Leibniz’s test could be used for z=-1. But my opinion- Apostol didn’t really give much attention to what seems a delicate matter, that is the radius of convergence for complex power series. Perhaps it should be obvious but I would imagine an appeal to De Moivre’s theorem with the rules for convergence of complex series would have been nice.
I think you could do something like let v = z/2. |v| = |z|/2 = 1, apply thm 10.19 on v to say v is bounded (when v isn’t equal to 1), note v^n = z^n/2^n, and use Dirichlet’s test from there
Thinking about this a little more, I think you’d have to justify that v^n = z^n/2^n a little more (since this isn’t always the case). (z_1z_2)^w = z_1^w*z_2^w when -pi < argz_1 + argz_2 < pi (if our restrictions on the argument are (-pi, pi]) (see 9.10 exercise 11). arg(1/2) = 0, and arg(z) in (-pi, pi], so v^n = z^n/2^n is valid