I wonder how to find a tight and simple upper bound for
$$\sum_{k=1}^N \frac 1 {ak^n - b}$$
What I thought to use was an integration comparison test
\begin{align} \sum_{k=1}^{N} \frac 1 {ak^n - b} &\leq \int_{1}^{N} \frac 1 {a(x-1)^2 - b} dx \\ &= \left[ - \frac{\ln \left|\sqrt{\frac ab} (x-1) + 1\right| - \ln \left|\sqrt{\frac ab} (x-1) - 1\right|}{2\sqrt {ab}} \right]_{1}^N \\ &= - \frac{\ln \left|\sqrt{\frac ab} (N-1) + 1\right| - \ln \left|\sqrt{\frac ab} (N-1) - 1\right|}{2\sqrt {ab}} \end{align}
If $n=2$. I find it hard to further simplify this expression for a polynomial form. Is there a better or simple way of developing the bound?
If you want to use $$\sum_{k=1}^{N} \frac 1 {k^n - c} \leq \int_{1}^{N} \frac {dx} {(x-1)^n - c} =I $$ the integral is given in terms of the Gaussian hypergeometric function $$I=\frac {N-1}c \,\,\, _2F_1\left(1,\frac{1}{n};1+\frac{1}{n};\frac{(N-1)^n}{c}\right)$$
But I think that you could compute the summations writing $$k^n-c=\prod_{m=1}^n(k-r_m)$$ Using partial fraction decomposition $$\frac 1{k^n-c}=\sum_{m=1}^n \frac {A_m}{k-r_m}$$ and using generalized harmonic numbers $$\sum_{k=1}^{N} \frac 1 {k - r_m}=H_{N-r_m}-H_{-r_m}$$