I know that $\sum_{k=n}^\infty \frac{1}{k^2} < \frac{1}{n-1} = O(\frac{1}{n})$. I confronted a similar situation but now I have some finite constant $c \in \mathbb{R}$ in the denominator: \begin{equation} \sum_{k=n}^\infty \frac{1}{k(k - c)}. \end{equation}
Can we upper bound this in terms of $n$ (and $c$ if needed) (using Big-O notation is completely fine)?
One thing I've tried is first expressing $\frac{1}{k(k-c)} = \frac{1}{c}(\frac{1}{k-c} - \frac{1}{k})$, so that our equation becomes
\begin{equation} \frac{1}{c}\sum_{k=n}^\infty (\frac{1}{k-c} - \frac{1}{k}). \end{equation} From here, I am stuck. I think some cancellations might help us, or maybe thinking about the absolute value ($\lvert \sum_{k=n}^\infty \frac{1}{k(k - c)} \rvert$), but I'm not sure. Any help would be much appreciated.
Here is a simple, explicit bound. For $n > \max\{0, c\}$, we have
\begin{align*} \sum_{k=n}^{\infty} \frac{1}{k(k - c)} &= \frac{1}{c} \sum_{k=n}^{\infty} \left( \frac{1}{k-c} - \frac{1}{k}\right) \\ &= \frac{1}{c} \sum_{k=n}^{\infty} \int_{0}^{1} x^{k-1}(x^{-c} - 1) \, \mathrm{d}x \\ &= \frac{1}{c} \int_{0}^{1} x^{n-1}\frac{x^{-c} - 1}{1-x} \, \mathrm{d}x \end{align*}
Now for $0 < x < 1$, we have
\begin{align*} \frac{x^{-c} - 1}{c} &= \int_{x}^{1} t^{-c-1} \, \mathrm{d}t \leq \begin{cases} x^{-c-1}(1-x), & c > -1, \\ 1-x, & c \leq -1. \end{cases} \end{align*}
Plugging this back to the last integral, for $n > \max\{0, c+1\}$ we obtain
\begin{align*} \sum_{k=n}^{\infty} \frac{1}{k(k - c)} \leq \frac{1}{n-\max\{0, c+1\}} \end{align*}
Or, if you are willing to include the digamma function into your vocabulary of closed-form expressions, we can compute the sum as
$$ \sum_{k=n}^{\infty} \frac{1}{k(k - c)} = \frac{\psi(n) - \psi(n-c)}{c} $$
and utilize asymptotic expansion of $\psi(\cdot)$ for further use.