Asymptotic development of a recurrent sequence Let $u_0 = 1$ and $u_{n+1} = \frac{u_n}{1+u_n^2}$ for all $n \in \mathbb{N}$.
I can show that $u_n \sim \frac{1}{\sqrt{2n}}$, but I would like one more term in the asymptotic development, something like $u_n = \frac{1}{\sqrt{2n}}+\frac{\alpha}{n\sqrt{n}} + o\bigl(\frac{1}{n^{3/2}}\bigr)$.
Here is the outline of my proof of $u_n \sim \frac{1}{\sqrt{2n}}$: 


*

*$(u_n)$ is decreasing, and bounded from below by $0$, hence converges.

*The limit $\ell$ satisifies $\ell = \frac{\ell}{1+\ell^2}$, hence $\ell = 0$.

*A computation gives $v_n = u_{n+1}^{-2} - u_n^{-2} \to 2$.

*Using Cesàro lemma, $\frac{1}{n} \sum_{k=0}^{n-1} v_k \to 2$.

*Hence $\frac{1}{n} u_n^{-2} \to 2$.


Let $a_n=\dfrac1{u_n^2}$. Then
$$
a_{n+1}-a_n = \left(\frac{1+u_n^2}{u_n}\right)^2 - \frac1{u_n^2} = 2+u_n^2 = 2+\frac1{a_n}.
$$
From this and $a_2=4$ we can find a successive sequence of estimates for $n\ge2$:
\begin{align*}
 a_n &\ge 2n; \\
 a_n &\le 2n + \sum_{k=2}^{n-1}\frac1{2k}
      < 2n+\frac12\log n; \\
 a_n &\ge 2n + \sum_{k=2}^{n-1}\frac1{2k+\frac12\log k}
      > 2n+\int_2^n \frac{dx}{2x+\frac12\log x}
      \\ &> 2n+\int_2^n \frac{dx}{2x}
            - \int_2^n \frac{\frac12\log x}{2x(2x+\frac12\log x)}dx
      = 2n+\frac12\log n - \mathcal{O}(1).
\end{align*}
Then 
$$
u_n = a_n^{-1/2} =
\frac1{\sqrt{2n}} \left(1+\frac{\log n}{4n}+\mathcal{O}\bigg(\frac1n\bigg)\right)^{-1/2} \\
= \frac1{\sqrt{2n}} \left(1-\frac12 \cdot \frac{\log n}{4n} + \mathcal{O}\bigg(\frac1n\bigg) \right) =
\frac1{\sqrt{2n}} - \frac1{8\sqrt2}\cdot \frac{\log n}{n^{3/2}} + \mathcal{O}\left(\frac1{n^{3/2}}\right).
$$
