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

You have $$\begin{aligned} a_{n+1} &=\sqrt{na_n +2n+1}\\ &= \sqrt{n^2+2n+1+(n a_n -n^2)}\\ &=(n+1) \sqrt{1+\left(\frac{n}{n+1}\right)^2\left(\frac{a_n}{n}-1\right)} \end{aligned}$$

Hence denoting $b_n =\frac{a_n}{n}$

$$\begin{aligned} b_{n+1} -1 &=\sqrt{1+\left(\frac{n}{n+1}\right)^2\left(b_n-1\right)}-1 \end{aligned}$$

As for $x\gt 0$

$$\begin{aligned} \sqrt{1+x}-1 &=\frac{x}{\sqrt{1+x}+1}\\ &\le \frac{1}{2}x \end{aligned}$$

you get

$$\begin{aligned}0 &\le b_{n+1}-1\\ &\le \frac{1}{2} \left(\frac{n}{n+1}\right)^2 (b_n-1)\\ &\le \frac{1}{2}(b_n-1) \end{aligned}$$ and can conclude to the desired result.

You can prove by induction on $n$ that $$ n \le a_n \le n + a_1 . $$ Indeed, for $n=1$, these inequalities hold. To move from $n$ to $n+1$, note that $$ a_{n + 1} \ge \sqrt {n \cdot n + 2n + 1} = \sqrt {(n + 1)^2 } = n + 1 $$ and \begin{align*} a_{n + 1} & \le \sqrt {n(n + a_1 ) + 2n + 1} = \sqrt {(n + 1)^2 + na_1 } \\ & \le \sqrt {(n + 1)^2 + 2(n + 1)a_1 + a_1^2 } = n + 1 + a_1 . \end{align*}

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 $b_{n+1}-1 \rightarrow 0$.