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For $x_n = \frac{1}{2+x_{n-1}}$ where $x_1 =1/2$, show that the sequence is monotonic and find its limit.

What I first did was finding $x_{n+1}$, which equals $\frac{1}{2+x_n}$; then $x_{n+2}=\frac{1}{2+\frac{1}{2+x_n}}=\frac{x_n+2}{2x_n+5}$ thus it does eventually get smaller hence $x_n>x_{n+1}$. How can I finish this?

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up vote 3 down vote accepted

The single most useful thing to do here is to draw on a same picture the graphs of the functions $u:x\mapsto1/(2+x)$ and $v:x\mapsto x$, say for $x$ in $(0,1)$. The rest follows by inspection...

Since $u$ is decreasing from $u(0)\gt0$ to $u(\infty)=0$, $u$ has a unique fixed point, say $x^*$. Since $u(x_1)\lt x_1$, one knows that $x_1\gt x^*$. Drawing on our picture the segments from $(x_1,0)$ to $(x_1,x_2)$ to $(x_2,x_2)$ to $(x_2,x_3)$ and so on, one sees that:

  • the sequence $(x_{2n-1})$ is decreasing and $x_{2n-1}\gt x^*$ for every $n$,
  • the sequence $(x_{2n})$ is increasing and $x_{2n}\lt x^*$ for every $n$,
  • the whole sequence $(x_{n})$ is neither decreasing nor increasing,
  • and the whole sequence $(x_{n})$ converges to $x^*$.

Numerically, $x^*=\sqrt2-1\approx.414$.

Edit: To show the last item, call $L$ the limit of $(x_{2n})$ and $M$ the limit of $(x_{2n-1})$, then $M=u(L)$ and $L=u(M)$ hence $L=u\circ u(L)$ and $M=u\circ u(M)$. Computing $u\circ u$, one sees that $x=u\circ u(x)$ is equivalent to $x=(2+x)/(5+2x)$, that is, $x^2+2x=1$, that is, $x=\pm\sqrt2+1$. Since $L$ and $M$ are positive, this shows that $L=M=x^*$ hence $(x_n)$ converges.

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Thank you very much! – Tom Sep 23 '13 at 21:03
@Tom Perhaps you should delete your now obsolete comments. – Vishal Gupta Sep 28 '13 at 4:59

We prove by induction that the odd subsequence is decreasing and that the even subsequence is increasing.

Proceed by observing that $x_{1} > x_{3}$ and $x_{2} < x_{4}$. Now assume that $$x_{2n-1} > x_{2n+1}, \quad x_{2n} < x_{2n+2}$$

Now we have

$$ x_{2n+3} = \frac{1}{2 + x_{2n+2}} < \frac{1}{2 + x_{2n}} = x_{2n+1}$$

With this, we have

$$ x_{2n+2} = \frac{1}{2 + x_{2n+1}} < \frac{1}{2 + x_{2n+3}} = x_{2n+4} $$

Thus, by induction we have our claim.

Also, prove by induction that $x_{2n} < x_{1}$ and $x_{2n-1} > x_{2}$ for all $n \in \mathbb{N}$.

Thus both the even and the odd subsequence converge. We need to show that they converge to the same value.

Now assume that the limit exists (justified below) and is equal to $L$. Then

$$L = \lim x_{n} = \lim\frac{1}{2 + x_{n-1}} = \frac{1}{2 + L}$$

from where we get that

$$ L = (\pm\sqrt{2} - 1)$$

But $L > 0$ (Why?) and hence $L = \sqrt{2} - 1$.

EDIT: For the existence of the limit, see the Edit in Did's proof.

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Thank you very much! – Tom Sep 23 '13 at 21:03

$ x_\infty=\frac{1}{2+x_\infty} \Rightarrow x_\infty^2+2x_\infty-1=0 \Rightarrow x_\infty= \sqrt2-1$ as summation of positive terms can not be negative

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@DavidH Perhaps you should delete your now obsolete comments. – Vishal Gupta Sep 28 '13 at 5:00

Hint: In the limit, you have $x = \frac{1}{2+x} $ or $x^2 + 2x - 1 = 0$.

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This iteration defines a continued fraction. Look at the continued fraction expansion of $\sqrt{2}$, subtract one from it and you have the continued fraction that you define in your post.

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