<s>[INST] Convergence of $\sum_{n=1}^\infty \frac{(-1)^n3^n}{2n+1}z^{2n+1}$ for $|z|= \frac{\sqrt{3}}{3}$ I have shown that the series $$\sum_{n=1}^\infty \frac{(-1)^n3^n}{2n+1}z^{2n+1}$$
converges for $|z|<\frac{\sqrt{3}}{3}$. Can anyone help me to study the convergence for $|z|= \frac{\sqrt{3}}{3}$?
 [/INST] We will use the series
$$
\arctan(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}
$$
Using Dirichlet's Test, we get that, except for $z=\pm i\frac{\sqrt3}3$, the series converges. Using Abel's Theorem, we get that for $z=\frac{\sqrt3}3e^{i\theta}$,
$$
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^n3^n}{2n+1}\left(\frac{\sqrt3}3e^{i\theta}\right)^{2n+1}
&=\frac{\sqrt3}3\sum_{n=1}^\infty(-1)^n\frac{e^{i\theta(2n+1)}}{2n+1}\\
&=\frac{\sqrt3}3\left(\arctan\left(e^{i\theta}\right)-1\right)\\
&=\frac{\sqrt3}3\left(\frac i2\log\left(\frac{i+e^{i\theta}}{i-e^{i\theta}}\right)-1\right)
\end{align}
$$
which only blows up when $e^{i\theta}=\pm i$.

For $z=\pm i\frac{\sqrt3}3$, we get
$$
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^n3^n}{2n+1}\left(\pm i\frac{\sqrt3}3\right)^{2n+1}
&=\pm i\frac{\sqrt3}3\sum_{n=1}^\infty\frac1{2n+1}
\end{align}
$$
which diverges by comparison to the Harmonic Series.
</s>