# 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 : $$u_n = \sqrt{n+u_{n-1}}= \sqrt{n}(1+\dfrac{1}{2\sqrt{n}} + o(1)) = \sqrt{n} + \dfrac{1}{2} + o(1)$$

• It will be much easier to help you if you show us what you've done so far. – Nick Peterson Sep 8 '17 at 20:12
• I added some info – MiKiDe Sep 8 '17 at 20:17

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: $$\boxed{u_n = \sqrt{n}+\frac{1}{2}+\frac{1}{8\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)}$$

Write $y_n = u_n - \sqrt{n} - \frac{1}{2}$. Then the recurrence becomes

$$y_n = \sqrt{n + \sqrt{n} + \frac{1}{2} + y_{n-1}} - \sqrt{n + \sqrt{n} + \frac{1}{4}} = \frac{\frac{1}{4} + y_{n-1}}{\sqrt{n + \sqrt{n} + \frac{1}{2} + y_{n-1}} + \sqrt{n} + \frac{1}{2}}.$$

Now look at $8\sqrt{n}\cdot y_n$ and use that you already know $y_n \to 0$.