Proving that a sequence $[(a_{(m,n)}) = \frac{v_n}{n},\; n \ge 1$] converges to $\frac{\sqrt m}{m}$, for any $m \gt 0$. 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.
 A: 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 $(2)$
$$\lim\limits_{n\rightarrow\infty}\frac{v_n}{n}=\frac{\sqrt{m}}{m}$$
A: 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.
