Let $t<0$ and $f(k)\in O(|k|^{t})$ a real function, $k\in\mathbb{Z}$. We consider $$a_n\cdot \sum_{k=1}^n \frac{1}{n} \frac{f(k)}{n^t}$$ where $a_n\subset \mathbb{R}$ and $a_n\xrightarrow{n\rightarrow\infty}0$.
We know that $$\sum_{k=1}^n \frac{1}{n} \frac{k^t}{n^t}\xrightarrow{n\rightarrow\infty}\int_0^1 x^t dx<\infty$$ and therefore $$a_n\cdot\sum_{k=1}^n \frac{1}{n} \frac{k^t}{n^t}\xrightarrow{n\rightarrow\infty}0.$$ Can we follow now that $$a_n\cdot \sum_{k=1}^n \frac{1}{n} \frac{f(k)}{n^t}\xrightarrow{n\rightarrow\infty}0 ?$$ I think it should hold, but I would be thankful for a short explanation. Thanks!
We know that $|f(k)|\le C\,|k|^t$ for some constant $C>0$. Then $$ \Bigl|\sum_{k=1}^n \frac{1}{n} \frac{f(k)}{n^t}\Bigr|\le C\sum_{k=1}^n \frac{1}{n} \frac{k^t}{n^t}. $$ The right hand side converges, so it is bounded, and so is the left hand side (although we do not know if it converges.) Since $a_n\to0$,the product also converges to $0$.