# Radius of Convergence of $\sum n^{-1}z^{3n}$

I want to find the Convergence Radius of the series $\sum_{n=1}^\infty n^{-1}z^{3n}$ ($z\in\mathbb{C}$). Firstly, i set $b_k := \begin{cases} a_n=1/k, & \text{for } k=3n, \ n\in \mathbb{N}^* \\ 0, & \text{else} \end{cases}$

Now I have $\sum_{k=1}^\infty b_kz^k$ and use Cauchy-Hadamard Criterion and get

$R=\frac{1}{\limsup |b_k|^{1/k}}=\frac{1}{\limsup \sqrt[3n]{|a_n|}}$

I know that $1/\limsup |a_k|^{1/n}=1$ and want to determine $R$ by the squeezing lemma.

This is where I am not quite sure anymore.

Can I make the estimate $\limsup |a_k|^{1/(3n)}\ge\limsup |a_k|^{1/n}=1$?

Because $|a_k|\le 1$ for all $k\in\mathbb{N}^*$ we should have $1\ge \limsup |a_k|^{1/(3n)}$ too, right?

I concluded $R=1$ but I have my doubts, as I am a total newbe in series.

• When $|z|<1$, the series converges, since$$\left|\frac{z^{3n}}n\right|=\frac{|z^3|^n}n\leqslant|z^3|^n$$and the series $\sum_{n=1}^\infty|z^3|^n$ converges (since $|z^3|=|z|^3<1$). Therefore, the radius of convergence is at least $1$.
• Since the series diverges when $z=1$, the radius of convergens is at most $1$.
Therefore, the radius of convergence is $1$.