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

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?

  • 6
    $\begingroup$ This seems absolutely correct to me. $\endgroup$ – TheSilverDoe Feb 28 at 9:55
  • $\begingroup$ I think The final answer is true. $\endgroup$ – Darman Feb 28 at 21:59

Your proof is correct. Here is another way to compute this limit.
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. $n\leq \frac{1}{x}<n+1$ or $n=\left\lfloor \frac{1}{x} \right\rfloor$ then $\forall i\in\mathbb{N}$: $$i\cdot n \leq \frac{i}{x}<i\cdot(n+1) \Rightarrow i\cdot n \leq \left\lfloor \frac{i}{x} \right\rfloor<i\cdot(n+1) \tag{1}$$ 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 $n\leq \frac{1}{x}<n+1 \Rightarrow \frac{1}{n}\geq x \geq \frac{1}{n+1}$ 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 $(3)$.


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