Find equivalent of recurrent sequence $u_1=1$, $u_{n+1}=1+\frac n {u_n}$ [duplicate]

The sequence is defined: $$u_1=1, u_{n+1}=1+\dfrac n {u_n}$$

The question asks to find an asymptotic development of 2 terms for $n\to+\infty$. I have got $u_n\sim_\infty\sqrt{n}$, but how to derive the 2nd term?

marked as duplicate by Did, Winther, Daniel W. Farlow, ronno, Arnaud D.Mar 5 '17 at 18:15

Let $u_{n}=\sqrt{n}+a+\dfrac{b}{\sqrt{n}}+O\left( \dfrac{1}{n} \right)$, then
\left \{ \begin{align*} a &= 1-a \\ b+\frac{1}{2} &= a^2-b \end{align*} \right. \implies (a,b)=\left( \frac{1}{2},-\frac{1}{8} \right)
• You've got $u_{n}\sim \sqrt{n}$, right? So $u_{n+1} \sim 1+\dfrac{n}{\sqrt{n}}=\sqrt{n}+O(1)$. It's natural to try the orders of every halves. – Ng Chung Tak Mar 2 '17 at 10:55