Determine if the following series converges or diverges. I'm struggling with this series. Could you try to help me me?

Determine if the following series converges or diverges.$$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\left(\frac{1+i\sqrt{3}}{2}\right)^n$$
  I think that I have to use Abel's test and I know the convergence of the first part of the equation. But I don't know how to deal with the second.

 A: First note that we have 
$$\begin{align}
\left(\frac{1+i\sqrt 3}{2}\right)^n&=\left(\cos(\pi/3)+i\sin(\pi/3)\right)^n\\\\
&=\cos(n\pi/3)+i\sin(n\pi/3)
\end{align}$$
Next, we can write
$$\sum_{n=1}^N \frac{\left(\frac{1+i\sqrt 3}{2}\right)^n}{\sqrt n}=\sum_{n=1}^N \frac{\cos(n\pi/3)}{\sqrt n}+i\sum_{n=1}^N \frac{\sin(n\pi/3)}{\sqrt n}$$
Note that $\frac{1}{\sqrt n}$ goes to zeo monotonically as $n\to \infty$ and that the partial sums, $\sum_{n=1}^N \cos(n\pi/3)$ and $\sum_{n=1}^N \sin(n\pi/3)$, are bounded as 
$$\left|\sum_{n=1}^N \cos(n\pi/3)\right|=\left|\sin\left((2N+1)\pi/6 \right)-\frac12\right|\le \frac32$$
and 
$$\left|\sum_{n=1}^N \cos(n\pi/3)\right|=\left|\frac{\sqrt 3}{2}-\cos\left((2N+1)\pi/6 \right)\right|\le \frac{\sqrt 3}2+1$$
Hence, Dirichlet's Test guarantees that both series $\sum_{n=1}^\infty \frac{\cos(n\pi/3)}{\sqrt n}$ and $\sum_{n=1}^\infty \frac{\sin(n\pi/3)}{\sqrt n}$ converge.

Therefore, the series of interest convereges.

A: Abel's test 
Be $\,x\in\mathbb{R}\,$ with $\,\displaystyle\frac{x}{2\pi}\notin \mathbb{Z}\,$ .
$\displaystyle |\sum\limits_{n=1}^\infty \frac{e^{ixn}}{\sqrt{n}}|= \frac{1}{|1-e^{ix}|}|e^{ix}-\sum\limits_{n=1}^\infty e^{ix(n+1)}(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})|$
$\displaystyle\hspace{2cm}\leq \frac{1}{\sqrt{2-2\cos x}}\left(1+\sum\limits_{n=1}^\infty (\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})\right)=\sqrt{\frac{2}{1-\cos x}}$ 
$\displaystyle \left(\frac{e^{ixn}}{\sqrt{n}}\right)_n\,$ is a null sequence therefore the (limited) series (above) is convergent.
With $\,\displaystyle x:=\frac{\pi}{3}\,$ and $\,\displaystyle e^{i\frac{\pi}{3}}=\frac{1}{2}+i\frac{\sqrt{3}}{2}\,$ we get $\,\displaystyle |\sum\limits_{n=1}^\infty \frac{(\frac{1}{2}+i\frac{\sqrt{3}}{2})^n}{\sqrt{n}}|\leq 2 \,$ .
