Third term of the asymptotic expansion of $u_n$ when $u_{n+1} = \sin(u_n)$ Lastly, I've been working on a problem :
Let $u_0\in \mathbb{R}$ and $u_{n+1} = \sin(u_n)$ for all $n\in\mathbb{N}$. Show that :
$$u_n = \sqrt{\frac{3}{n}} - \frac{3\sqrt{3}}{10}\frac{\ln(n)}{n\sqrt{n}} + o\left(\frac{\ln(n)}{n\sqrt{n}}\right)$$
I managed to prove it, but I was wondering whether or not it would be possible to express any third term in this expansion?
Thanks for your help!
 A: In de Bruijn's classic
"Asymptotic Methods in Analysis",
chapter 8,
it is shown (8.6.5) that
$$x_n
=\sin_n(x_0)
=\sqrt{3/n}\left(1-\dfrac{3}{10}\dfrac{\log(n)}{n}-\dfrac{C}{n}
+\dfrac{a\log^2(n)+b\log(n)+c}{n^2}+O\left(\dfrac{\log^3(n)}{n^3}\right)\right)
$$
where
$C$ depends on $x_0$ and
$$a = \dfrac{27}{209},
b = \dfrac{9}{20}C,
c = \dfrac38 C^2-\dfrac{3}{10}C+\dfrac{79}{200}.
$$
A: Did you use the series for $\csc^2x$ to find a recursion for $u^{−2}_n$? The $O(1)$ term of $u^{−2}_n$ will have $1/u^2_0$ and the $\gamma$ constant, and they become $O(1/n\sqrt n)$ in $u_n$.
Oops, I think there is also a whole stream of $\zeta(k)$ terms, all in $O(n^{-3/2})$
Edit:
$$\csc^2x=x^{-2}+\frac13+\frac{x^2}{15}+\frac{2x^4}{189}+\ldots\\
u_{n+1}^{-2}=u_n^{-2}+\frac13+\frac{u_n^2}{15}+\ldots\\
u_n^{-2}=\frac n3+u_0^{-2}\ldots$$
But then, when you feed that back, you get $(\gamma+\ln n)/5$, then $(2/189)\sum(9/n^2)$ and so on.  The $\ln n$ term is what you found, but there is a lot of O(1) terms in $u_n^{-2}$, giving an $O(n^{-3/2})$ term in $u_n$
