Convergence/divergence of a couple of infinite series I have to figure out if the following series converge or diverge:

(1) $\sum_{n=1}^{\infty}\frac{(-i)^{n+1}}{n^{2}+1}$. This has the $N$-th partial sum $s_N=\sum^{N}\frac{(-i)^{N+1}}{N^{2}+1}$.

My attempt: To show whether or not the series converges, I need to show whether {$S_{N}$} is Cauchy. So I have to see that $\forall\epsilon>0$ $\exists N_{\epsilon}$ s.t. for all $n,m>N_{\epsilon}$, $|\sum_{k=m+1}^{n}\frac{(-i)^{n+1}}{n^{2}+1}|<\epsilon$. Where do I go from here?

(2) $\sum_{n=1}^{\infty}e^{in\theta}/n$ for $0<\theta<2\pi$, $\theta$ fixed.

Not sure how to approach this. 
 A: When you try to determine the convergence/divergence of a series, always remember that there is not only one trick to do it. So if your attempt failed it can either mean that you didn't do it right or that maybe it's just not a right way to go.
For the first one ; using comparison, you can show that it is absolutely convergent since
$$
\sum_{n=1}^{\infty} \left| \frac{(-i)^{n+1}}{n^2+1} \right| = \sum_{n=1}^{\infty} \frac 1{n^2+1} \le \sum_{n=1}^{\infty} \frac 1{n^2} 
$$
and you know that the latter is a convergent series, hence your series converges absolutely.
For the second one, to know what you're looking for, you can informally do this :
\begin{align}
\frac 1i \sum_{n=1}^{\infty} \frac{e^{in\theta}}{n} = \sum_{n=1}^{\infty} \frac{e^{in\theta}}{in} = \sum_{n=1}^{\infty} \int e^{in\theta} \, d \theta = \int \sum_{n=1}^{\infty} (e^{i\theta})^n \, d\theta = \int \frac 1{1-e^{i\theta}} \, d\theta 
\end{align}
and from here you can compute this integral : by letting $u=1-e^{i\theta}$, you have
\begin{align}
\int \frac 1{1-e^{i\theta}}  d\theta & = \int \frac{i}{u(1-u)}  du = -i \int \frac 1{u(u-1)}  du = -i \int \left(  \frac 1{u-1} - \frac 1u \right) \\ du \\\
& = -i \log(u-1) + i \log(u) \\\
& = -i \log(-e^{i\theta}) + i \log(1-e^{i\theta}) \\\
& = i \log \left( \frac{1-e^{i\theta}}{-e^{i \theta}} \right) \\\
& = i \log(1 - e^{-i\theta}).
\end{align}
This made me noticed what I wanted ; the function we're looking for is a logarithm. If you compute the power series for $\log$, you get
$$
\log(1-z) = \sum_{n=1}^{\infty} \frac {z^n}{n}
$$
which means that your series sums to $-\log(1-e^{-i\theta})$, as long as $\theta \neq 0$ (or an integer multiple of $2\pi$). The trickier part comes because $|e^{i\theta}| = 1$, which means it is on the boundary of the disc of convergence of the function $\log(1-z)$. I must say that at the moment I don't know how to deal with it, but I'll try thinking a little more.
Hope that helps,
