$$\sum_{n=1}^\infty \frac{i^n}{n}$$
I'm not sure how to approach this series to find if it converges/diverges because of the $i^n$. I tried using the comparison test comparing it with $\frac{1}{n}$ and the ratio test, but didn't get anywhere. How would you approach this and look at series of this form or how to look at $i^n$ in an infinite series?
On
Since $i^n = i^{n+4}$, each 4 consecutive terms is of the form
$\begin{array}\\ \frac{i}{4m+1}+\frac{-1}{4m+2}+\frac{-i}{4m+3}+\frac{1}{4m+4} &=\frac{i}{4m+1}+\frac{-1}{4m+2}+\frac{-i}{4m+3}+\frac{1}{4m+4}\\ &=i(\frac{1}{4m+1}-\frac{1}{4m+3})-\frac{1}{4m+2}+\frac{1}{4m+4}\\ &=i(\frac{(4m+3)-(4m+1)}{(4m+1)(4m+3)})-(\frac{(4m+4)-(4m+2)}{(4m+2)(4m+4)})\\ &=i(\frac{2}{(4m+1)(4m+3)})-(\frac{2}{(4m+2)(4m+4)})\\ \end{array} $
Since each of the real and imaginary parts are less than $\frac{2}{(4m+1)^2}$, and the sum of these converges absolutely, the overall sum converges conditionally.
To write this out explicitly, this shows that
$\begin{array}\\ |\sum_{k=1}^{4n} \dfrac{i^k}{k}| &=|\sum_{m=1}^n (i(\dfrac{2}{(4m+1)(4m+3)})-(\dfrac{2}{(4m+2)(4m+4)})|\\ &\le\sum_{m=1}^n |(\dfrac{2}{(4m+1)(4m+3)})|+\sum_{m=1}^n|(\dfrac{2}{(4m+2)(4m+4)})|\\ &\lt\sum_{m=1}^n |(\dfrac{2}{(4m+1)^2})|+\sum_{m=1}^n|(\dfrac{2}{(4m+2)^2})|\\ &=\sum_{m=1}^n 2(\dfrac{1}{(4m+1)^2}+\dfrac{1}{(4m+2)^2})\\ \end{array} $
and this sum converges.
There are at most 3 additional terms in any sum beyond the $4n$ and these contribute at most $\dfrac{3}{4n}$ which goes to zero, so they do not affect the convergence.
$\sum_{n=1}^\infty\frac{i^n}n=\sum_{n=1}^\infty\frac{i^{2n}}{2n}+ \sum_{n=1}^\infty\frac{i^{2n-1}}{2n-1}=\sum_{n=1}^\infty\frac{(i^2)^{n}}{2n}+ i\sum_{n=1}^\infty\frac{(i^2)^{n-1}}{2n-1}$$=\sum_{n=1}^\infty\frac{(-1)^{n}}{2n}+ i\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2n-1}$
Now, you have two alternating series. Use the Alternating Series Test to see if they converge.
A similar idea goes into how we derive the relation $e^{ix}=\cos x +i\sin x$.