Convergence of complex series $sum_{n=1}^{infty}frac{i^n}{n}$

Prove that the series $\displaystyle \sum_{n=1}^{\infty}\frac{i^n}{n}$ converges.

Optional. find it's sum, if possible.

Comments. I am aware of the general result about the convergence (not absolute)
of $\displaystyle \sum_{n=1}^{\infty}\frac{z^n}{n}$, for $z\neq 1$. But i feel that the answer to the above problem can be more trivial, although have not made any interesting approach so far.

Thank you in advance!

Answer

We have $\frac{1}{1-z}=\sum_{n\geq 0}z^n$ for any $z\in\mathbb{C}$ such that $|z|<1$. The convergence is uniform over any compact subset of $\{z\in\mathbb{C}:|z|<1\}$, hence we are allowed to state

$$ \int_{0}^{i}\frac{dz}{1-z} = \sum_{n\geq 0}\frac{i^{n+1}}{n+1} = \sum_{n\geq 1}\frac{i^n}{n} $$
where the LHS equals
$$ -\log(1-i) = -\log\left(\sqrt{2}\, e^{-\frac{\pi i}{4}}\right) = \color{red}{-\frac{\log 2}{2}+\frac{\pi i}{4}}.$$

The convergence of the original series is granted by Dirichlet's test, as already remarked by other users. You may also notice that the real part is given by the terms of the original series with $n\in 2\mathbb{N}$ and the imaginary part is given by the terms of the original series with $n\in 2\mathbb{N}+1$. On the other hand,
$$ \sum_{n\geq 1}\frac{(-1)^n}{n}=-\log(2),\qquad \sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\frac{\pi}{4} $$
are pretty well-known. They can be proved with the same approach, restricted to the real line only.