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.

$$\begin{align}\sum_{n=1}^\infty\frac{(i)^n}{n}&=\sum_{n=1}^\infty\frac{(i)^{2n}}{2n}+\sum_{n=0}^\infty\frac{(i)^{2n+1}}{2n+1} \\ & =\sum_{n=1}^\infty\frac{(-1)^n}{2n}+i\sum_{n=0}^\infty\frac{(-1)^n}{2n+1} \\ \end{align}$$
Now apply Leibnitz test for alternating series.
