Help with proving that this series diverges I have this complex series $\sum\limits_{n=1}^{\infty} \frac{i^n}{\sqrt{n}}$, which I'm trying to prove converges.
Now, I know that a complex sequence converges iff both its real and its imaginary partial sums converge.
Obviously, $\sum\limits_{n=1}^{\infty} \Im(a_n)=\sum\limits_{n=1}^{\infty}0 \to 0$.
But, when I try to compute $\sum\limits_{n=1}^{\infty} \Re(a_n),$ I end up with $\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n}},$ which, whilst, intuitively, I know diverges, I can't seem to prove.
I know I have to use the Comparison Test, but to what should I compare it?
By the way, 'my' version of the Comparison Test is as follows (for divergence):
If $\frac{b_n}{a_n}$ is bounded above and if $\sum\limits_{n=1}^{\infty} b_n$ diverges, then $\sum\limits_{n=1}^{\infty}a_n$ diverges.
Could someone give me a hint?
Thanks
 A: 
Obviously, $\sum\limits_{n=1}^{\infty} \Im(a_n)=\sum\limits_{n=1}^{\infty}0 \to 0$.

This is not the case.
As  André Nicolas suggests in the comments, it is best to write out the first few terms to get a feel for the series:
$$
\frac{i}{\sqrt1}
 + \frac{-1}{\sqrt{2}}
 + \frac{-i}{\sqrt3}
 + \frac{1}{\sqrt4}
 + \frac{i}{\sqrt5}
 + \frac{-1}{\sqrt6}
 + \frac{-i}{\sqrt7}
 + \frac{1}{\sqrt9}
 + \frac{i}{\sqrt{10}}
 + \cdots
$$
You correctly observe that you should look at the real and imaginary parts.
\begin{align*}
\sum \Im (a_n) &= \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{5}} - \frac{1}{\sqrt{7}} + \cdots \\
\sum \Re (a_n) &= -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{4}} - \frac{1}{\sqrt{6}} + \frac{1}{\sqrt{8}} - \cdots \\
\end{align*}
The proper way to show these converge is with the alternating series test.

If you were trying to prove that the real and/or imaginary partial sums diverge (by the Alternating Series Test), could you just take the odd subsequence and show that that diverges?

The alternating series test requires these two conditions:


*

*The terms alternate in sign.

*The terms decrease in absolute value.
The fact that every other term diverges does not help you.
Note: The comparison test in the form it is usually stated says that if a complex or real series is bounded in absolute value by a positive convergent series, then that series converges (absolutely).  You cannot apply this form of the comparison test to the above series, because the sum of the absolute values of the terms diverges (the series is conditionally convergent).
