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I am trying to get an asymptotic expansion of the following sequence: $u_0 = 1$ and $u_{n+1} = \arctan u_n$. Here is what I tried:

First I noticed that $u_n \rightarrow 0$, and a limited expansion with Cesaro theorem gave me $u_n ^{-2} \sim \frac{2n}{3}$, i.e. $u_n \sim \sqrt{\frac{3}{2n}}$

I am trying to push the asymptotic expansion further: $u_{n+1}^{-2} - u_n^{-2} = \frac{2}{3} - \frac{1}{15}u_n^2 + o(u_n^2)$. Therefore, with a telescopic sum : $u_n^{-2} - u_0^{-2} = \frac{2n}{3} - \frac{1}{15} \sum{u_k^2} + \sum{o(u_k^2)}$. Since $u_k^2 \sim \frac{3}{2n}$, the comparison relations summation theorem gives us : $\sum{u_k^2} \sim \frac{3}{2}\sum{\frac{1}{k}} = \frac{3}{2}\sum{\frac{1}{k}} + o(\sum{\frac{1}{k}}) = \frac{3}{2}\ln n + o(\ln n)$ (Am I right until here?)

The same argument gives $\sum{o(u_k^2)} = o(\ln n)$.

So $u_n^{-2} = \frac{2n}{3} - \frac{1}{10}\ln n + o(\ln n)$.

Another limited expansion finally gives: $u_n = \sqrt{\frac{3}{2n}} + \frac{3}{40}\sqrt{\frac{3}{2}}\frac{\ln n }{n\sqrt{n}} + o\left(\frac{\ln{n}}{n\sqrt{n}}\right)$.

Is this asymptotic expansion right?

Thank you very much for your help,

AF

Arthur Filippi
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    just a reference, not an answer: the general case of iterative sequences where the function has a power series $x-a_kx^k + \cdots$ for $k>1$ and $a_k >0$ is treated in de Bruijn, Asymptotic Methods in Analysis chapter 8. Your leading term is definitely right. These two one two are closely related to your question. – Matthew Towers Oct 31 '23 at 11:37
  • $u_n^{-2} =\color{red}{1}+ \frac{2n}{3} - \frac{1}{10}\ln n + o(\ln n)$ would be better – Claude Leibovici Nov 02 '23 at 08:30
  • @ClaudeLeibovici yes but $1 = o(ln(n))$ – Arthur Filippi Nov 02 '23 at 13:09

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