This is an application of the Dirichlet's test also termed the generalized alternating test. The test goes as follows. If we have a sequence of positive real numbers, $\{a_n\}_{n=1}^{\infty}$ and a sequence of complex numbers $\{b_n\}_{n=1}^{\infty}$ such that
- $a_n$ is a decreasing sequence i.e. $a_n > a_m$ whenever $m>n$.
- $\lim_{n \rightarrow \infty} a_n = 0$.
- $\left \lvert \displaystyle \sum_{k=0}^{n} b_k\right \rvert \leq M$, $\forall n \in \mathbb{Z}^+$, where $M$ is some constant independent of $n$
then the series $\displaystyle\sum_{n=0}^{\infty} a_n b_n$ converges.
The proof is by partial summation technique. Let $S_n = \displaystyle \sum_{k=0}^n a_k b_k$ and $B_n = \displaystyle \sum_{k=0}^n b_k$.
By partial summation, we have that $\displaystyle S_n = a_{n+1} B_n + \sum_{k=0}^n B_k (a_k - a_{k+1})$. Hence, $\displaystyle \lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} \left( a_{n+1} B_n + \sum_{k=0}^n B_k (a_k - a_{k+1}) \right) = 0 + \sum_{k=0}^{\infty} B_k (a_k - a_{k+1})$.
Hence, all we need to show now is that $\displaystyle \sum_{k=0}^{\infty} B_k (a_k - a_{k+1})$ converges.
We will show that $\displaystyle \sum_{k=0}^{\infty} B_k (a_k - a_{k+1})$ in fact converges absolutely. Note that
\begin{align} \left \lvert \sum_{k=0}^n B_k (a_k - a_{k+1}) \right \rvert & \leq \sum_{k=0}^n \left \lvert B_k (a_k - a_{k+1}) \right \rvert\\ & = \sum_{k=0}^n \left \lvert B_k \right \rvert \left \lvert (a_k - a_{k+1}) \right \rvert\\ & \leq M \sum_{k=0}^{n} \left( a_k - a_{k+1}\right)\\ & \leq M(a_0 - a_{n+1})\\ & \leq M a_0 \end{align}
Hence, we have that $\displaystyle \lim_{n \rightarrow \infty} S_n$ exists.
For your problem, $a_n = \dfrac1n$ and $b_n = \exp(in \theta)$. Clearly, $a_n$ and $b_n$ satisfy the assumptions for the Dirichlet test to be applicable. The only thing which might not be clear is why $\left \lvert \displaystyle \sum_{k=1}^{n}b_k \right \rvert$ bounded. But as Sam has shown $\left \lvert \sum_{k=1}^{n}b_k \right \rvert = \left \lvert \dfrac{\exp(in \theta) - \exp(i (n+1) \theta)}{1 - \exp(i \theta)} \right \rvert \leq \dfrac{2}{\left \lvert 1 - \exp(i \theta) \right \rvert}$ which is some finite real number independent of $n$. (Note that $\theta \neq 2 k \pi$).
Hence, $\displaystyle \sum_{k=1}^{\infty} \dfrac{z^{k}}{k}$ converges for $z$ s.t. $|z| = 1$, except for $z=1$