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.

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.
