The harmonic progression has a simple and elegant formula, though it's arguably not a closed-form. If $a$ and $b$ are integers:
$$\sum_{j=1}^{n}\frac{1}{a j+b}=-\frac{1}{2b}+\frac{1}{2(a n+b)}-\pi\int_{0}^{1}(1-u)(\cos{2\pi(a n+b)u}-\cos{2\pi b u})\cot{\pi a u}\,du $$
We can generalize this formula, which in the case of odd powers is given by:
\begin{multline}
\sum_{j=1}^{n}\frac{1}{(a j+b)^{2k+1}}=-\frac{1}{2b^{2k+1}}+\frac{1}{2(a n+b)^{2k+1}}+\\
-\frac{(-1)^{k}(2\pi)^{2k+1}}{2}\int_{0}^{1}\sum_{j=0}^{k}\frac{B_{2j}\left(2-2^{2j}\right)(1-u)^{2k+1-2j}}{(2j)!(2k+1-2j)!}\left(\cos{2\pi(a n+b)u}-\cos{2\pi bu}\right)\cot{\pi au}\,du
\end{multline}
Since there is more than one way to derive these formulas, another possibility is below ($i$ is the imaginary unit):
\begin{multline}
\sum_{j=1}^{n}\frac{1}{(a j+b)^{k}}=-\frac{1}{2b^{k}}+\frac{1}{2(a n+b)^{k}}\\+\frac{i(2\pi\,i)^{k}}{2}\int_{0}^{1}\sum_{j=0}^{k}\frac{B_{j}(1-u)^{k-j}}{j!(k-j)!}\left(e^{2\pi i(a n+b)u}-e^{2\pi i\,bu}\right)\cot{\pi a u}\,du
\end{multline}
The demonstration on how one goes about deriving these formulas was given in this paper.