Cfrgtkky

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.

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

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

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.