<s>[INST] Let $a_{n+1}=\sqrt{a_n+n}$ with $a_1=1$ then find $\lim_{n \to \infty} (a_n -\sqrt{n})$ 
Let $a_{n+1}=\sqrt{a_n+n}$ with $a_1=1$ then find 
  $\displaystyle \lim_{n \to \infty} (a_n -\sqrt{n}).$

My Try:
$$\lim_n( \sqrt{a_{n}+n}-\sqrt{n}) \times \dfrac{( \sqrt{a_{n}+n}+\sqrt{n}) }{( \sqrt{a_{n}+n}+\sqrt{n})}=\lim_n \dfrac{1}{( \sqrt{\dfrac{1}{a_{n}}+\dfrac{n}{a_{n}^2}}+\sqrt{\dfrac{n}{a_{n}^2}})}.$$
Now what do I do?
 [/INST] We first show by induction that for any $n\geq 1$,
$$\sqrt{n}\leq a_n\leq \sqrt{n+2\sqrt{n}}\tag{1}.$$
The inequalities hold for $n=1$. As regards the inductive step we have that for $n\geq 1$, 
$$a_{n+1}=\sqrt{a_n+n}\geq \sqrt{\sqrt{n}+n}\geq \sqrt{1+n},$$
and 
$$a_{n+1}=\sqrt{a_n+n}\leq \sqrt{\sqrt{n+2\sqrt{n}}+n}
\leq \sqrt{(\sqrt{n}+1)+n}\leq \sqrt{2\sqrt{n+1}+n+1}.$$
Hence, by (1) and by the Squeeze Theorem,
$$1\leq \frac{a_n}{\sqrt{n}}\leq \sqrt{1+\frac{2}{\sqrt{n}}}\to 1\implies
\lim_{n\to \infty}\frac{a_n}{\sqrt{n}}=1$$
Finally, as $n$ goes to infinity,
$$
\begin{align}
a_{n+1} -\sqrt{n+1}&=\sqrt{a_n+n} -\sqrt{n+1}=\frac{a_n-1}{\sqrt{a_n+n} +\sqrt{n+1}}\\&=\frac{\frac{a_n}{\sqrt{n}}-\frac{1}{\sqrt{n}}}{\sqrt{\frac{a_n}{n}+1} +\sqrt{1+\frac{1}{n}}}\to \frac{1}{2}.
\end{align}$$
</s>