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$\forall n\in\mathbb{N}^*, u_n = \sqrt{n+u_{n-1}}$ 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 : $u_n = \sqrt{n+u_{n-1}}= \sqrt{n}(1+\dfrac{1}{2\sqrt{n}} + o(1)) = \sqrt{n} + \dfrac{1}{2} + o(1)$

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  • $\begingroup$ It will be much easier to help you if you show us what you've done so far. $\endgroup$ – Nick Peterson Sep 8 '17 at 20:12
  • $\begingroup$ I added some info $\endgroup$ – MiKiDe Sep 8 '17 at 20:17
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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: $$ \boxed{u_n = \sqrt{n}+\frac{1}{2}+\frac{1}{8\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)} $$

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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 $y_n \to 0$.

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