Compute the set of convergence of $\sum_{n=1}^{\infty}\frac{(-1)^n3^nz^{2n+1}}{2n+1}$

Question

Compute the set of convergence of the following sum $$\sum_{n=1}^{\infty}\frac{(-1)^n3^nz^{2n+1}}{2n+1}$$

Ok, so I defined my

$$ a_k = \begin{cases} \frac{(-1)^n3^n}{2n+1} & \quad \text{if } k=2n+1\\ 0 & \quad \text{otherwise } \\ \end{cases} $$

So I can calculate easily my radius of convergence as

$$\limsup|a_k|^{1/k}=\limsup \left| \frac{(-1)^n3^n}{2n+1} \right|^{1/n}=3^{1/2}$$

And therefore $\rho = 1/\sqrt{3}$

Everything cool until now.. When I get a $z \in \Bbb{C}$ such that $|z|=1/\sqrt{3}$ I don't know how to manipulate my series as to bound my partial sums and use Dirichlet's Criteria! What should I do!

Answer 1

Did you consider that perhaps it does not converge for some points on the boundary of the disk of convergence? Try $z = \frac{i}{\sqrt{3}}$ so that the signs cancel out. And try to find other such points of divergence. $\def\less{\smallsetminus}$

If you ignore the scalings, your function is just $z \mapsto \ln(1+iz)-\ln(1-iz)$ within the region $D[0,1] \less \{i,-i\}$, by Taylor series of $\ln$ and its convergence behaviour.

You may also be interested in Examples of Taylor series with interesting convergence along the boundary of convergence?