<s>[INST] Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that $a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$ 
Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that
$$a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$$

Let $b_{n}= \dfrac{a_{n}}{n},$ we need to prove $\lim b_{n}= 1,$ then we find a value $\beta\in\left ( 0, 1 \right )$ satisfying $\left | b_{n+ 1}- 1 \right |\leq\beta\left | b_{n}- 1 \right |$ so that
$$\beta\rightarrow 0\,{\rm aut}\,n\rightarrow\infty$$
I can't predict the relationship among them, I need the the help to go to the induction, thank you.
 [/INST] You don't need $\beta \rightarrow 0$, you just need that for sufficiently large $n$ $$\beta < 1$$ Then $$|b_{n+M}-1|<\beta^n|b_M-1| \rightarrow 0$$
You can check that
$$ b_{n+1} = \sqrt{\frac{b_nn^2 +2n+1}{(n+1)^2}}{} = \sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1}$$
that is
$$ b_{n+1}-1 = \sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1} - 1= \frac{(b_n-1)\frac{n^2}{(n+1)^2}}{\sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1} +1} $$
Since $a_1 \ge 1$, then $b_1 -1 \ge 0$ and by induction it's easy to show that $b_n -1 \ge 0$. We have then
$$ \frac{\frac{n^2}{(n+1)^2}}{\sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1} +1} \le  \frac{\frac{n^2}{(n+1)^2}}{\sqrt{1}+1} = \frac{n^2}{2(n+1)^2} \le \frac12 $$
Therefore
$$b_{n+1} -1  \le \frac12(b_n-1)$$
As showed before, it is enough to prove that $b_{n+1}-1 \rightarrow 0$.
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