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

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.

HINT: Notice that $$\sum_{n=1}^\infty \frac{i^n}{n}=\sum_{n=1}^\infty \frac{(-1)^n}{2n}-i\sum_{n=1}^\infty \frac{(-1)^n}{2n-1}$$

Hint: Look at ther partial sums is equal to:

$$\frac{i}{1}-\frac{1}{2}-\frac{i}{3}+\frac{1}{4}+\ldots \frac{i^n}{n}$$

Factor the complex part and the imaginary part. Notice that you have alternating series in the real and imaginary part. Both series converge by the Leibniz criterion. Additionally compare the real and imaginary part with the Taylor series of $\arctan(x)$ and $\ln(1+x)$.