Is the series $\sum_{n=1}^\infty \frac{(i)^n}{n}$ convergent or divergent Does the series
$$\sum_{n=1}^\infty\frac{(i)^n}{n}$$
converge or diverge? The only theorems I've covered thus far are:
Theorem: A series $\sum_{n=1}^\infty (x_n+i y_n)$ converges if and only if $\sum_{n=1}^\infty x_n$ and $\sum_{n=1}^\infty y_n$ converge.
Theorem: If $\sum_{n=1}^\infty z_n$ and $\sum_{n=1}^\infty w_n$ converge, then $\lim_{n\to\infty}z_n=0$.
Theorem: If $\sum_{n=1}^\infty z_n$ and $\sum_{n=1}^\infty w_n$ converge, then
$$\sum_{n=1}^\infty cz_n=c\sum_{n=1}^\infty z_n$$
and 
$$\sum_{n=1}^\infty(z_n+w_n)=\sum_{n=1}^\infty z_n+\sum_{n=1}^\infty w_n$$
Theorem: The comparison test.
Thanks.

We can write
$$\begin{align}
\sum_{n=1}^{2N} \frac{i^n}{n}&=\sum_{n=1}^{2N} \frac{\left(e^{i\pi/2}\right)^n}{n}\\\\
&=\underbrace{\sum_{n=1}^{2N} \frac{\cos(n\pi/2)}{n}}_{\text{All of the odd terms are zero}}+i \underbrace{\sum_{n=1}^{2N} \frac{\sin(n\pi/2)}{n}}_{\text{All of the even terms are zero}}\\\\
&=\sum_{n=1}^{N} \frac{(-1)^n}{2n}+i \sum_{n=1}^{N} \frac{(-1)^{n-1}}{2n-1}\\\\
\end{align}$$
Now apply Leibniz's Test.
