# Find $\lim\limits_{x\to0^+}x(\lfloor \frac{1}{x}\rfloor+\lfloor \frac{2}{x}\rfloor+\cdots+\lfloor \frac{k}{x}\rfloor), \, k \in \mathbb N$.

$$M:=x\left(\left\lfloor \frac{1}{x} \right\rfloor+\left\lfloor \frac{2}{x}\right\rfloor+\cdots+\left\lfloor \frac{k}{x}\right\rfloor\right),\, k \in \mathbb N.$$

Using $$\lfloor y \rfloor=y-\{y\}$$,

$$M=x\sum_{i=1}^{k}\left(\frac{i}{x}-\left\{\frac{i}{x}\right\}\right)=\frac{k(k+1)}{2}-x \sum_{i=1}^{k}\left\{\frac{i}{x}\right\}$$

Since $$0\leq\{y\}<1$$, the coefficient of $$x$$ is finite and thus, $$\lim_{x\to0^+}M=\frac{k(k+1)}{2}$$

Is this correct?

• This seems absolutely correct to me. – TheSilverDoe Feb 28 at 9:55
• I think The final answer is true. – Darman Feb 28 at 21:59

Note that $$\frac{i}{x}-1 < \left \lfloor \frac{i}{x} \right \rfloor \le \frac{i}{x}$$,$$\forall i=\overline{1,k}$$.
Therefore, $$\frac{k(k+1)}{2x} -k < \left \lfloor \frac{1}{x} \right \rfloor + \left \lfloor \frac{2}{x} \right \rfloor +...+ \left \lfloor \frac{k}{x} \right \rfloor \le \frac{k(k+1)}{2x}$$.
Hence, $$\frac{k(k+1)}{2} -kx < x\left(\left \lfloor \frac{1}{x} \right \rfloor + \left \lfloor \frac{2}{x} \right \rfloor +...+ \left \lfloor \frac{k}{x} \right \rfloor \right) \le \frac{k(k+1)}{2}$$.
By the squeeze theorem we get that your limit equals $$\frac{k(k+1)}{2}$$ as you proved.
Another way is, because $$x$$ is approaching $$0^+$$, i.e. it becomes very small and is positive, then $$\exists n\in \mathbb{N}$$ s.t. $$n\leq \frac{1}{x} or $$n=\left\lfloor \frac{1}{x} \right\rfloor$$ then $$\forall i\in\mathbb{N}$$: $$i\cdot n \leq \frac{i}{x} and $$n\left(\sum\limits_{i=1}^k i\right)\leq\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor <(n+1)\left(\sum\limits_{i=1}^k i\right) \iff \\ n\cdot \frac{k(k+1)}{2}\leq\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor <(n+1)\cdot \frac{k(k+1)}{2} \tag{2}$$ but $$n\leq \frac{1}{x} and $$(2)$$ becomes $$\frac{n}{n+1}\cdot \frac{k(k+1)}{2} < x\cdot n\cdot \frac{k(k+1)}{2}\leq \\ x\left(\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor\right) < \\ x\cdot(n+1)\cdot \frac{k(k+1)}{2} \leq \frac{(n+1)}{n}\cdot \frac{k(k+1)}{2}$$ or $$\frac{n}{n+1}\cdot \frac{k(k+1)}{2} < x\left(\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor\right) < \frac{(n+1)}{n}\cdot \frac{k(k+1)}{2} \tag{3}$$ Now, $$\lim\limits_{x\rightarrow0^+} \equiv \lim\limits_{n\rightarrow\infty}$$ and the result follows from $$(3)$$.