Proving that a sequence $[(a_{(m,n)}) = \frac{v_n}{n},\; n \ge 1$] converges to $\frac{\sqrt m}{m}$, for any $m \gt 0$.

Question

Given the natural numbers with $0 \in \mathbb N$.

Let $m \in \mathbb N$ with $m \ge 1$.

We define the sequence $(a_{(m,n)})_{\,n \ge 1}$ in $\mathbb Q$ as follows:

$\quad \text{Set } u_n = [\frac{n \times n}{m}] \quad \text{ ([.] is the floor function)}$.

$\quad \text{Set } v_n = \text{ the greatest natural number with } v_n^2 \le u_n$.

$\quad \text{Set } a_{(m,n)} = \frac{v_n}{n}$.

Question: Is there a proof that the sequence $(a_{(m,n)})_{\,n \ge 0}$ converges to $\frac{\sqrt m}{m}?$

For the special case $m = 2$, see Proving that a sequence [$a_n = \frac{v_n}{n},\; n \ge 2$] converges.

Answer 1

Similarly with $m=2$ case, but using divisibility with remainder, given $n\in\mathbb{N}$ $$n=mq+r, 0\leq r <m \Rightarrow \left[\frac{n\cdot n}{m}\right]=q^2m+2qr+\left[\frac{r^2}{m}\right] \tag{1}$$ also, we have $$v_n \leq \sqrt{u_n} < v_n +1 \Rightarrow \frac{v_n}{n}\leq \frac{\sqrt{u_n}}{n}<\frac{v_n}{n} + \frac{1}{n} \overset{(1)}{\Rightarrow} \\ \frac{v_n}{n}\leq \frac{\sqrt{q^2m+2qr+\left[\frac{r^2}{m}\right]}}{mq+r}<\frac{v_n}{n} + \frac{1}{n}\Rightarrow \\ \frac{v_n}{n}\leq \frac{\sqrt{m}}{m}\cdot\frac{\sqrt{1+\frac{2r}{qm}+\frac{\left[\frac{r^2}{m}\right]}{q^2m}}}{1+\frac{r}{mq}}<\frac{v_n}{n} + \frac{1}{n}$$ or $$0\leq \frac{\sqrt{m}}{m}\cdot\frac{\sqrt{1+\frac{2r}{qm}+\frac{\left[\frac{r^2}{m}\right]}{q^2m}}}{1+\frac{r}{mq}} - \frac{v_n}{n} < \frac{1}{n} \tag{2}$$

But, from $(1)$, $m$ is a fixed number and $r$ is bounded, so when $n\rightarrow\infty$ then $q\rightarrow\infty$. Thus $$\lim\limits_{n\rightarrow\infty} \frac{\sqrt{1+\frac{2r}{qm}+\frac{\left[\frac{r^2}{m}\right]}{q^2m}}}{1+\frac{r}{mq}}= \lim\limits_{q\rightarrow\infty} \frac{\sqrt{1+\frac{2r}{qm}+\frac{\left[\frac{r^2}{m}\right]}{q^2m}}}{1+\frac{r}{mq}}=1$$ and finally from $(2)$ $$\lim\limits_{n\rightarrow\infty}\frac{v_n}{n}=\frac{\sqrt{m}}{m}$$

Answer 2

The integer part satisfies \begin{alignat*}{3} \frac{n^2}{m}-1 & \le u_n && \le \frac{n^2}{m} \\ \sqrt{u_n}-1 & \le v_n && \le \sqrt{u_n} \end{alignat*} Plugging this in, we get $$ \sqrt{\frac 1m - \frac{1}{n^2}} - \frac 1n %\left( \sqrt{\frac 1m - \frac{1}{n^2}} - 1 \right) / n \le \frac{v_n}{n} \le \frac{1}{\sqrt{m}}. $$ The right-hand side is already $1/\sqrt{m}$, and the left-hand side tends to this number.