We do know it's approximate value to be $\ln{k}+\gamma$ when $a=1$ and $b=1$. Can we have some similar formula to approximate this value where $a$, and $b$ are natural numbers?
PS: I've figured out the value when $a=4$ and $b=3$ or $b=1$ through the use of series of $\ln 2$, $\pi/4$ and the formula for approximate value of the harmonic numbers. Also, we can easily figure out it's value when $b=1$ and $a=2$ but what about the general case.
For simplicity, I shall assume $ab>0$. So $$S_k=\sum\limits_{n=0}^{k}\frac{1}{an+b}=\frac 1b+\sum\limits_{n=1}^{k}\frac{1}{an+b}=\frac 1b+\frac 1a\sum\limits_{n=1}^{k}\frac{1}{n+\frac ba}=\frac 1b+\frac 1a\sum\limits_{n=1}^{k}\frac{1}{n+c}$$ (using $c=\frac ba$ for more simplicity). So, $$S_k=\frac 1b+\frac 1a\left(H_{c+k}-H_c \right)$$ Now, using the results given in this paper, a very good approximation of the harmonic numbers is given by $$H_m\approx \log \left(m+\frac{1}{2}\right)+\gamma+\frac{5}{30 \left(2m+1\right)^2+{21}}$$ which is said to overestimate the result by $$\frac{2071}{12600\, (2 m+1)^6}$$ (have a look here) giving $$H_{c+k}-H_c \approx \log \left(\frac{2 (c+k)+1}{2 c+1}\right)+\frac{5}{30 (2 (c+k)+1)^2+21}-\frac{5}{30 (2 c+1)^2+21}$$ and then the approximation for $S_k$.
For illustration purposes, let us try using $a=\frac 12$, $b=\frac43$. This should give $$S_k\approx 2 \log \left(1+\frac{6 k}{19}\right)+\frac{10899}{14692}+\frac{30}{360 k^2+2280 k+3673}$$ and, for some values of $k$, the following table. $$\left( \begin{array}{ccc} k & \text{exact} & \text{approximation} \\ 100 & 7.709168268 & 7.709172706 \\ 200 & 9.064524171 & 9.064528609 \\ 300 & 9.865035157 & 9.865039595 \\ 400 & 10.43516943 & 10.43517387 \\ 500 & 10.87831208 & 10.87831652 \\ 600 & 11.24085617 & 11.24086061 \\ 700 & 11.54765689 & 11.54766132 \\ 800 & 11.81359346 & 11.81359789 \\ 900 & 12.04828315 & 12.04828758 \\ 1000 & 12.25830280 & 12.25830724 \end{array} \right)$$
