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I want to use this method to prove convergence of: $a_{n}=\sqrt{2-a_{n-1}}$, $a_{0}=\frac{2}{3}$.

Here is my attempt at proof:

It can be proven inductively that $0< a_{n}<2$ for all $n$.

I want to show that each of the odd and even subsequences are monotone.

$a_{2n+3}-a_{2n+1}=\sqrt{2-a_{2n+2}}-\sqrt{2-a_{2n}}=\frac{a_{2n}-a_{2n+2}}{\sqrt{2-a_{2n+2}}+\sqrt{2-a_{2n}}}=\frac{a_{2n}-\sqrt{2-\sqrt{2-a_{2n+1}}}}{\sqrt{2-a_{2n+2}}+\sqrt{2-a_{2n}}}$

I'm stuck as I'm not able to show that $(a_{2n+1})_{n\geq0}$ is decreasing.

Thank's in advance.

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    The odd-indexed subsequence is decreasing. The even-indexed subsequence is increasing. – 2017-01-12
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    @DanielFischer thank you! I edited. Can you please give me a hint as to how I can show that the odd-indexed is decreasing for example? – 2017-01-12
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    You have already $$a_{2n+3} - a_{2n+1} = \frac{-(a_{2n+2} - a_{2n})}{\text{positive stuff}},$$ and you get an analogous formula for the difference of two successive even-indexed terms. Then starting from $a_2 - a_0 > 0$ (which you must verify by hand), you can show that $(a_{2n})$ is increasing and $(a_{2n+1})$ is decreasing by induction. You can prove it simultaneously for both subsequences, or you can do one step more in the recurrence - ignoring all positive factors - to get $a_{k+2} - a_k = (\text{something positive})\cdot (a_k - a_{k-2})$ and treat each subsequence separately. – 2017-01-12

2 Answers 2

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$$ \begin{align} &\, a_{n}=\sqrt{2-a_{n-1}}\,\Rightarrow\,a_{n}\ge0 \\[2mm] &\, a_{n}^2=2-a_{n-1}\,\Rightarrow\,a_{n-1}=2-a_{n}^2=\left(\sqrt{2}-a_{n}\right)\left(\sqrt{2}+a_{n}\right)\,\ge0 \\[2mm] &\, \qquad\Rightarrow\,\sqrt{2}-a_{n}\ge0\,\Rightarrow\,a_{n}\le\sqrt{2} \\[8mm] &\, a_{n}-a_{n-2}=\left(2-a_{n+1}^2\right)-\left(2-a_{n-1}^2\right)=a_{n-1}^2-a_{n+1}^2=\left(a_{n-1}-a_{n+1}\right)\left(a_{n-1}+a_{n+1}\right) \\[2mm] &\, \qquad =\left[\left(2-a_{n}^2\right)-\left(2-a_{n+2}^2\right)\right]\left(a_{n-1}+a_{n+1}\right)=\left(a_{n+2}^2-a_{n}^2\right)\left(a_{n-1}+a_{n+1}\right) \\[2mm] &\, \qquad =\left(a_{n+2}-a_{n}\right)\left(a_{n+2}+a_{n}\right)\left(a_{n-1}+a_{n+1}\right)\,\Rightarrow \,\text{sgn}\left(a_{n+2}-a_{n}\right)=\text{sgn}\left(a_{n}-a_{n-2}\right) \\[2mm] &\, \qquad\Rightarrow\,\text{Either }\,\,\,\left\{\,a_{n+2}\ge a_{n}\ge a_{n-2}\,\right\}\,\,\text{ or }\,\,\left\{\,a_{n+2}\le a_{n}\le a_{n-2}\,\right\} \\[2mm] &\, \qquad\Rightarrow\,\text{Both }\quad\left\{a_{2n-1}\right\}\,\,\text{ and }\,\,\left\{a_{2n}\right\}\,\text{ are }\,\color{red}{\text{ monotonic}} \\[2mm] &\, \qquad\Rightarrow\begin{cases} a_0=\,\frac{2}{3}\,\,\approx0.667,\quad a_2\approx0.919\,\gt a_0 &\Rightarrow\color{red}{\left\{a_{2n}\right\}\,\text{ increasing}}\\[2mm] a_1=\frac{2}{\small\sqrt{3}}\approx1.155,\quad a_3\approx1.040\,\lt a_1 &\Rightarrow\color{red}{\left\{a_{2n-1}\right\}\,\text{ decreasing}} \end{cases} \\[2mm] &\, \qquad\text{And because}\,\left\{\,0\le a_{n}\le\sqrt{2}\,\right\}\,\Rightarrow\,\text{Both }\,\,\left\{a_{2n-1}\right\}\,\text{ and }\,\left\{a_{2n}\right\}\,\text{ are }\,\color{red}{\text{convergent}} \\[8mm] &\, \text{Let: }\,\,L=\lim_{n\rightarrow\infty}a_{n},\quad L_{o}=\lim_{n\rightarrow\infty}a_{2n-1},\quad L_{e}=\lim_{n\rightarrow\infty}a_{2n} \\[2mm] &\, \qquad\Rightarrow\,L_o=2-L_e^2\,\,\,{\small\&}\,\,\,L_e=2-L_o^2 \\[2mm] &\, \qquad\Rightarrow\,L_o-L_e=L_o^2-L_e^2=\left(L_o-L_e\right)\left(L_o+L_e\right) \\[2mm] &\, \qquad\Rightarrow\,L_o-L_e=0 \quad\text{ or }\quad L_o+L_e=1 \\[2mm] &\, L_o+L_e=1\,\Rightarrow\,\begin{cases} L_o^2=2-L_e=1+L_o &\Rightarrow\,L_o=\frac{1\pm\sqrt{5}}{2}\,\notin\left[0,\sqrt{2}\right]\\[2mm] L_e^2=2-L_o=1+L_e &\Rightarrow\,L_e=\frac{1\pm\sqrt{5}}{2}\,\notin\left[0,\sqrt{2}\right] \end{cases} \\[2mm] &\, L_o-L_e=0\,\Rightarrow\,L_o=L_e=L=\sqrt{2-L} \\[2mm] &\, \qquad\qquad\quad\,\,\Rightarrow\,L^2+L-2=0\,\Rightarrow\,L=\frac{-1\pm3}{2}\,\Rightarrow\,\color{red}{L=1} \\[8mm] &\, \text{Generally, }\quad \color{blue}{\lim_{n\rightarrow\infty}a_{2n-1}=\lim_{n\rightarrow\infty}a_{2n}}\color{red}{=\lim_{n\rightarrow\infty}a_{n}=1\,\,\colon\,\forall\,\,a_0\in[-2,\,+2]} \end{align} $$

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If $01-x/(2-x)$ (by squaring both sides and calculating). This deserves more publicity. We will use this in (2).(ii). below.

If $a_n$ converges to $L$ then $0\leq L=\sqrt {2-L}\implies (L^2=2-L\land 0\leq L)\implies L=1.$ So it is natural to let $a_n=1+b_n.$

We have $0<|b_n|<1.$ (Equivalent to $0

(1). If $1>b_n>0$ then $1+b_{n+1}=\sqrt {2-(1+b_n)}=\sqrt {1-b_n}<1$, which implies $b_{n+1}<0.$..... If $0>b_n>-1$ then $1+b_{n+1}=\sqrt {1-b_n}>1$, which implies $b_{n+1}>0.$

(2).(i). If $b_n<0$ then $b_{n+1}>0$ so $$1+2|b_{n+1}|=1+2b_{n+1}<(1+b_{n+1})^2=1-b_n=1+|b_n|$$ which implies $0<|b_{n+1}|<|b_n|/2.$

(2). (ii). If $b_n>0$ then $b_{n+1}<0$ so $$1-|b_{n+1}|=1+b_{n+1}=\sqrt {1-b_n}\;> 1-b_n/(2-b_n)$$ which implies $|b_{n+1}|

Applying all this, we have:

(3).(i). If $b_n<0$ we have $b_{n+1}>0$ so $|b_{n+2}|<|b_{n+1}|<|b_n|/2.$

(3).(ii). If $b_n>0$ we have $b_{n+1}<0$ so $|b_{n+2}|<|b_{n+1}|/2<|b_n|/2.$