Studying the convergence of $\sum_\limits{n=1}^{\infty}\frac{e^{2in}}{n\sqrt{n}}$ 
Study the convergence of the following series:$$\sum_\limits{n=1}^{\infty}\frac{e^{2in}}{n\sqrt{n}}$$

I thought of applying the root test:
$\lim_{n\to\infty}({\frac{e^{2in}}{n\sqrt{n}}})^{\frac{1}{n}}=\lim_{n\to\infty}\frac{e^{2i}}{n^{\frac{3}{2n}}}=e^i$
However is $e^i>1$?
Questions:
1) Is the application of the root test right? What conclusion can I take?
2) Which would be other alternative ways of solving the question?
Thanks in advance!
 A: When you apply the root test to a series $\displaystyle\sum_{n=1}^\infty a_n$, what you should compute is the limit $\displaystyle\lim_{n\to\infty}\sqrt[n]{\lvert a_n\rvert}$. That is not what you did.
On the other hand, $(\forall n\in\mathbb N):\left\lvert\dfrac{e^{2in}}{n\sqrt n}\right\rvert=\dfrac1{n^{3/2}}$. Since the series $\displaystyle\sum_{n=1}^\infty\dfrac1{n^{3/2}}$ converges (by the integral test), your series converges absolutely. In particular, it converges.
A: As an alternative, if you know that $\sum \frac{1}{n^c}$ converges for $c>1$, and that $e^{2in} = \cos(2n) + i\sin(2n)$ then this isn't too hard.
Consider the series $$\sum |\frac{\cos(2n)}{n√n}| < \sum \frac{1}{n^{\frac{3}{2}}},$$ and from here, we may conclude the original series converges. 
EDIT: additionally, to answer your question about $e^{i} > 1$: this isn't true, but it contrarily, is not true that $e^{i}<1$. $e^{i} = \cos(1) + i\sin(1)$, and you are trying to compare this to a strictly real number. It is tantamount to asking if $i$ is greater than $1$. There's no total ordering on the complex numbers, so these kinds of comparisons don't make sense. Instead, you could consider $|e^{i}|$ which is, in fact, equal to $1$. 
A: Application of the ROOT TEST shows 
$$\limsup_{n\to\infty}\sqrt[n]{\left|\frac{e^{i2n}}{n^{3/2}}\right|}=\lim_{n\to \infty}\sqrt[n]{n^{-3/2}}=1$$
So, the root test is inconclusive.  
On the other hand, Dirichlet's Test is applicable since $\frac{1}{n^{3/2}}$ monotonically decreases to $0$ and for all $N$, $\left|\sum_{n=1}^N e^{i2n} \right|$ is bounded (in fact it is bounded by $\csc(1)$).  
The power of Dirichlet's test is not of full display here since $\sum_{n=1}^\infty \frac{e^{i2n}}{n^{3/2}}$ converges absolutely.  But note that Dirichlet's test guarantees that the series $\sum_{n=1}^\infty \frac{e^{i2n}}{n^a}$ converges for all $a>0$!
