TITLE: Evaluating $\sum_{k=1}^{n} \lfloor{xk}\rfloor$
QUESTION [0 upvotes]: I don't know how to proceed with it. Even if i can get to an approximation that would also be fine.
Here $ 2 \lt x \lt 4 $ and $n \gt {10}^{10}$

REPLY [1 votes]: There exists the average fractional part of $kx$, which is between $0$ and $1$. Let this number $a$.$$\sum_{k=1}^{n} \lfloor{xk}\rfloor=\sum_{k=1}^{n} \left(xk-a\right)=\frac{xn(n+1)}{2}-an\approx\frac{xn(n+1)}{2}$$
Last approximation was possible because $\frac{x(n+1)}{2}$ is much larger than $1$, upper bound of $a$.
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