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I want to prove that $\frac{1}{\sqrt 2}\leq x_{n}\leq 1$ for all $n\in\mathbb N$ for the sequence $(x_n)_{n\in\mathbb N}$ define by $$x_{n+1}=\frac{x_n+1}{2x_n+1}$$ and $x_0=1$.

It works for $n=0$, then I suppose $\frac{1}{\sqrt 2}\leq x_n\leq 1$, and so $\frac{1}{2x_n+1}\geq \frac{1}{3}$ and $x_n+1\geq \frac{\sqrt 2+1}{\sqrt 2}$ and so

$$x_{n+1}\geq \frac{\sqrt 2+1}{3\sqrt 2}$$ but this is smaller than $\frac{1}{\sqrt 2}$ and I can't do better.

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  • $\begingroup$ Try to find the range of f(x)=(x+1)/(2x+1), $x \in [\frac{1){\sqrt2},1]$. $\endgroup$ – user175968 Oct 17 '14 at 15:26
  • $\begingroup$ Try to find the range of f(x)=(x+1)/(2x+1), $x \in [\frac{1){\sqrt2},1]$ $\endgroup$ – user175968 Oct 17 '14 at 15:26
  • $\begingroup$ Try to find the range of $f(x)=(x+1)/(2x+1)$, $x \in [\frac{1}{\sqrt2},1]$ $\endgroup$ – user175968 Oct 17 '14 at 15:28
  • $\begingroup$ $\frac{x_n+1}{2x_n+3}\geq \frac{1}{\sqrt 2}\Rightarrow \sqrt{2}x_n+\sqrt{2}\geq 2x_n+3$ i.e., $\sqrt{2}-3\geq x_n(2-\sqrt{2})$ which is not quite good as $x_{n}\geq 0$ for all $n$... so, some thing is not quite correct with the question i guess.. $\endgroup$ – user87543 Oct 17 '14 at 15:28
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    $\begingroup$ If $x_0 = 1$, then $x_1 = {2 \over 5} < {1 \over \sqrt{2}}$? (Am i missing something?) $\endgroup$ – copper.hat Oct 17 '14 at 15:29
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Let $u_n=\dfrac{\sqrt{2}x_n-1}{\sqrt{2}x_n+1}$. Then it is straightforward to check that $$ u_{n+1}=\frac{\sqrt{2} x_n+\sqrt{2}-2x_n-1}{\sqrt{2} x_n+\sqrt{2}+2x_n+1} =\frac{1-\sqrt{2}}{1+\sqrt{2}}u_n=\frac{-1}{(1+\sqrt{2})^2}u_n $$ This implies that $$ u_n=\frac{(-1)^n}{(1+\sqrt{2})^{2n}}u_0=\frac{(-1)^n}{(1+\sqrt{2})^{2n+2}} $$ So, $x_{2n}>\dfrac{1}{\sqrt{2}}$ and $x_{2n+1}<\dfrac{1}{\sqrt{2}}$.

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It is obvious that $0<x_n<1$.

Let's look on two domains - $x_n\in\left(0,\frac{1}{\sqrt{2}}\right]$ and $x_n\in\left(\frac{1}{\sqrt{2}},1\right]$.

For the first domain we will get that $x_{n+1}\in\left[\frac{1}{\sqrt{2}},1 \right)$ and for the second domain we will get $x_{n+1}\in\left(0,\frac{1}{\sqrt{2}}\right)$, hence the series alternates between these two ranges.

Assuming that the series converges, i.e $\displaystyle \exists L\in\mathbb{R} \ s.t \ \lim_{n\to\infty}x_n=L$, it is easy to find that $L=\frac{1}{\sqrt{2}}$.

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