$\{a_{i}\}_{i=1}^\infty$ is a sequence where $a_{1}=1$, $a_{n+1}=\frac{1}{2}(a_{n}+\frac{n}{a_{n}})$ Let $\{a_{i}\}_{i=1}^\infty$ be a sequence which has the properties
$a_{1}=1$,
$a_{n+1}=\frac{1}{2}(a_{n}+\frac{n}{a_{n}})$.
Find $\lceil a_{2022}\rceil$ and $\displaystyle\lim_{n\to\infty}\frac{a_{n}}{\sqrt{n}}$.
 A: Both statements can be achieved if we prove that $\sqrt{n-1} \leq a_n \leq \sqrt{n}$.
Since by the squeeze theorem  $$\lim_{n \to \infty} \frac{a_n}{\sqrt{n}} = 1$$
and for the $\lceil a_{2022}\rceil$ term:
$$ 44.955 \lt \sqrt{2021} \leq a_{2022} \leq \sqrt{2022} \lt 44.96$$
implies that $\lceil a_{2022}\rceil = 45$.
Proving the lowerbound
Since $a_n$ are all positive we can use the AM-GM in-equality $\frac{x+y}{2} \geq \sqrt{xy}$, with $x=a_n, y= n/a_n$.
This implies that $a_{n+1} = \frac{a_n + n/a_n}{2} \geq \sqrt{n}$, or $a_{n} \geq \sqrt{n-1}$
Proving the upperbound
This is equivalent to proving that $a_{n+1} \geq a_{n}$, since this implies $$ a_{n+1} a_n \geq a_n^2 $$
$$ a_{n+1} a_n - a_n^2 \geq 0 $$
$$ 2(a_{n+1} a_n - a_n^2) \geq 0 $$
$$ 2 a_{n+1} a_n - 2 a_n^2 \geq 0 $$
$$ a_n^2 + n - 2 a_n^2 \geq 0 $$
$$ n - a_n^2 \geq 0 $$
$$ n \geq a_n^2 $$
$$ a_n^2 \leq n $$
$$ a_n \leq \sqrt{n} $$
I will add the proof that this is an increasing sequence later, having trouble coming up with a good proof of it.
A: This is just to complete Emil Kerimov's answer,
and verify that $$ \sqrt{n} \le a_{n+1} \le \sqrt{n+1} \quad (*)$$ for all $n$.
The lower bound follows from   the AMGM inequality, as noted in Emil Kerimov's answer.
To verify the upper bound in $(*)$, we may assume that $n \ge 2$. The function
$f_n(x)=x+n/x$ is decreasing in $ (0,\sqrt{n}]$, since $f_n'(x)=1-n/x^2$.
Thus, using the lower bound $a_n \ge \sqrt{n-1}$, we get $$a_{n+1}=\frac{f_n(a_n)}{2} \le \frac{f_{n}( \sqrt{n-1})}{2} \,.$$
Therefore,
$$4a_{n+1}^2\le f_{n}( \sqrt{n-1})^2=n-1+2n+\frac{n^2}{n-1} < 4n+4 \,.$$
This proves the upper bound in $(*)$.
