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I am stuck on the following questions :

Let $(u_{n})_{n \in \mathbb{N}}$ the sequence defined by : $u_{0} > 0$, $u_{1} > 0$ and :

$$ \forall n \in \mathbb{N}, \; u_{n+2} = \sqrt{u_{n+1}}+\sqrt{u_{n}} $$

I have already proved that : $\forall n \in \mathbb{N}, \; 0 < m \leq u_{n} \leq M$, where $m = \min(u_{0},u_{1},4)$ and $M=\max(u_{0},u_{1},4)$. Let's assume that the sequence $(u_{n})_{n \in \mathbb{N}}$ converges to $\ell$. I proved that the limit $\ell$ is 4. (In fact, it can be either $0$ or $4$ but if the sequence converges to $0$, there would exist some $N \in \mathbb{N}$ such that : $\forall n \in \mathbb{N}, \; n \geq N, \; u_{n} < \frac{m}{2}$, which is impossible.)

For all $n \in \mathbb{N}$, let $\alpha_{n} = \vert u_{n} - \ell \vert$. I would like to prove that there exists $k \in ]0,\frac{1}{2}[$ such that :

$$ \forall n \in \mathbb{N}, \; \alpha_{n+2} \leq k(\alpha_{n+1} + \alpha_{n}) $$

Here is what I tried :

Let $n \in \mathbb{N}$.

$$ \begin{eqnarray*} \alpha_{n+2} & = & \vert u_{n+2} - 4 \vert \\ & = & \vert \sqrt{u_{n+1}} + \sqrt{u_{n}} - 4 \vert \\ & \leq & \vert \sqrt{u_{n+1}} - 2 \vert + \vert \sqrt{u_{n}} - 2 \vert \\ \end{eqnarray*} $$

The derivative of $x \, \mapsto \, \sqrt{x}$ on $[m,M]$ is such that : $\forall x \in [m,M], \; \frac{1}{2\sqrt{x}} \leq \frac{1}{2\sqrt{m}}$. So, it follows that :

$$ \forall (x,y) \in [m,M]^{2}, \; \vert \sqrt{x}-\sqrt{y} \vert \leq \frac{1}{2\sqrt{m}} \vert x - y \vert $$

So (since $4 \in [m,M]$), I get :

$$ \alpha_{n+2} \leq \frac{1}{2\sqrt{m}} (\alpha_{n}+\alpha_{n+1}) $$

Is that correct ? Thanks for your help!

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up vote 1 down vote accepted

This all looks fine and shows good ideas and effort. Let me mention a different trick based on the binomial formulas that often helps with square roots without resorting to derivatives: $$|\sqrt x-\sqrt y|=\frac{|x-y|}{\sqrt x+\sqrt y}\le\frac1{2\sqrt{\min\{x,y\}}} \cdot|x-y|$$ But note that either way $m$ may be too small to make $k=\frac1{2\sqrt m}$ work.

Instead, I'd try the more direct estimate $$\begin{align} \alpha_{n+2}&\le|\sqrt{u_{n+1}}-2|+|\sqrt{u_n}-2|\\&=\frac{\alpha_{n+1}}{\sqrt{u_{n+1}}+2}+\frac{\alpha_{n}}{\sqrt{u_{n}}+2}\\ &\le\frac{\alpha_{n+1}+\alpha_n}{2+\sqrt m}.\end{align}$$ Therefore with $\beta_n=\max\{\alpha_{2n},\alpha_{2n+1}\}$ and $\kappa=\frac2{2+\sqrt m}<1$ $$ \alpha_{2n+2}\le\kappa\beta_n,\qquad\alpha_{2n+3}\le \frac{\kappa\beta_n+\beta_n}{2+\sqrt m}\le \kappa\beta_n,$$ hence $$\beta_{n+1}\le \kappa\beta_n$$ so that $\beta_n\to 0$, $\alpha_n\to 0$, $u_n\to 4$.

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