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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?

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

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Let $u_{n}=\sqrt{n}+a+\dfrac{b}{\sqrt{n}}+O\left( \dfrac{1}{n} \right)$, then

\begin{align*} u_{n+1} &\sim 1+\frac{n}{\sqrt{n}+a+\frac{b}{\sqrt{n}}} \\ \sqrt{n+1}+a+\frac{b}{\sqrt{n+1}} &\sim 1+\frac{n}{\sqrt{n}+a+\frac{b}{\sqrt{n}}} \\ \sqrt{n}+a+\frac{b+\frac{1}{2}}{\sqrt{n}} &\sim \sqrt{n}+(1-a)+\frac{a^2-b}{\sqrt{n}} \\ \end{align*}

Equating corresponding orders:

$$ \left \{ \begin{align*} a &= 1-a \\ b+\frac{1}{2} &= a^2-b \end{align*} \right. \implies (a,b)=\left( \frac{1}{2},-\frac{1}{8} \right)$$

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  • $\begingroup$ But why the power differs by 1/2 every time? $\endgroup$ – pqros Mar 1 '17 at 23:31
  • $\begingroup$ 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. $\endgroup$ – Ng Chung Tak Mar 2 '17 at 10:55
  • $\begingroup$ This approach can prove nothing, at most it can help to check that if such an expansion holds then its coefficients are such and such. What about attacking the question itself? $\endgroup$ – Did Mar 5 '17 at 12:52

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