Show that for $n\to \infty, u_n - \sqrt{n} - \frac{1}{2} \sim \frac{1}{8\sqrt{n}}$ $\forall n\in\mathbb{N}^*, u_n = \sqrt{n+u_{n-1}}$ with $u_1 = 1$.
I've already shown that $ u_n \sim \sqrt{n}$ and that $\underset{n\rightarrow \infty}{\lim} u_n - \sqrt{n} = \dfrac{1}{2}$
How to show that $n\rightarrow \infty, u_n - \sqrt{n} - \dfrac{1}{2} \sim \dfrac{1}{8\sqrt{n}}$ ?  (Or it might not be this, maybe something else ? I dont really know... I intuited this result with the asymptotic exampsion of $\sqrt{n}$.
I know that I have to use the asymptotic expansion, but I don't know how to use it correctly (I've some difficulties to keep the $o(1)$).
What I've done :
$u_n = \sqrt{n+u_{n-1}}= \sqrt{n}(1+\dfrac{1}{2\sqrt{n}} + o(1)) = \sqrt{n} + \dfrac{1}{2} + o(1)$

Letting $u_n = \sqrt{n}+\frac{1}{2}+\varepsilon_n$ with $\lim_{n\to\infty} \varepsilon_n=0$, you have
$$
\sqrt{n}+\frac{1}{2}+\varepsilon_n = \sqrt{n+\sqrt{n}+\frac{1}{2}+\varepsilon_{n-1}}
$$
for all $n$. Doing a Taylor expansion of the RHS yields, using the fact that $\varepsilon_{n-1}=o(1)$:
$$\begin{align}
\sqrt{n}+\frac{1}{2}+\varepsilon_n &= \sqrt{n}\sqrt{1+\frac{1}{\sqrt{n}}+\frac{1}{2n}+o\left(\frac{1}{n}\right)}
\\
&= \sqrt{n}\left(1+\frac{1}{2\sqrt{n}}+\frac{1}{4n}+o\left(\frac{1}{n}\right)-\frac{1}{8n}\right)
\\
&= \sqrt{n}+\frac{1}{2}+\frac{1}{8\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)
\end{align}$$
and so $\varepsilon_n = \frac{1}{8\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)$, giving the result:
$$
\boxed{u_n = \sqrt{n}+\frac{1}{2}+\frac{1}{8\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)}
$$
