$\sum_{n=1}^\infty \frac{i^n}{n}$ converge? $$\sum_{n=1}^\infty \frac{i^n}{n}$$
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?
 A: $\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 $e^{ix}=\cos x +i\sin x$.
A: 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.
