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Define $\{a_n\}$ by $a_1=1/2, a_{n+1}=\sqrt{1-a_n}$. Does it converge?

My Work:

This sequence is neither increasing nor decreasing, but intuitively i can see that it converges. So in such case how do I show that it converges? Please help.

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  • $\begingroup$ Consider the monotonicity of the subsequences formed by the terms with even and odd indexes. Then look at the distance between consecutive terms. $\endgroup$ – Pp.. Jan 3 '15 at 7:24
  • $\begingroup$ Using Excel, I see it looks like it converges to 0.618033989 after around 85 terms. $\endgroup$ – Clarinetist Jan 3 '15 at 7:28
  • $\begingroup$ @ Jihad $\sqrt{5}$, not $\sqrt{3}$ $\endgroup$ – Extremal Jan 3 '15 at 7:32
  • $\begingroup$ @Mathi, yep $\frac{-1+\sqrt{5}}{2}$. $\endgroup$ – Jihad Jan 3 '15 at 7:32
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Let $f(x) = \sqrt{1-x}$. If you can find an interval $[a,b]$ such that $f([a,b]) \subseteq [a,b]$ and that for some $k \le 1$, you have $|f'(x)| \le k$ for all $x \in [a,b]$, then $f(x) = \sqrt{1-x}$ is a contraction map on $[a,b]$.

This means there is a unique $x_0 \in [a,b]$ such that $f(x_0) = x_0$, and the sequence $\{a_n\}$ given by $a_{n+1} = f(a_n)$ converges to $x_0$ for any initial term $a_1 \in [a,b]$.

It's not hard to pick an interval like $[\frac{1}{2},\frac{1}{\sqrt{2}}]$ and show that the above conditions are satisfied.

Also, you can find this fixed point by solving $x_0 = \sqrt{1-x_0}$.

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define the function $$ f:(0,1) \to (1,\frac54] $$ by the transformation $$ f(x) = x+\sqrt{1-x} $$ the terms of the series satisfy ($n \ge 1$) $$ a_{n+1}^2 - a_n^2 = a_n-a_{n-1} $$ giving $$ \frac{|a_{n+1}-a_{n}|}{|a_n - a_{n-1}|} = \frac1{a_{n+1}+a_n} =\frac1{f(a_n)}\lt 1 $$ this may be useful, though further work is required

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    $\begingroup$ The trouble with this clever trick is that it is... a clever trick, while entirely standardized approaches exist, which are probably what the OP should concentrate on (say, the content of @JimmyK4542's answer). $\endgroup$ – Did Jan 3 '15 at 9:49
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First of all, it is easy to check that $0\leq a_n\leq 1$.

Now, by evaluating $a_{n+1}-a_n$, you can find that $a_{n+1}\leq a_n$ iff $a_n\geq \frac{-1+\sqrt5}{2}$, and by the definision you can find that $a_{n+1}\geq \frac{-1+\sqrt5}{2}$ iff $a_n\leq \frac{-1+\sqrt5}{2}$.

Finally, you can prove by induction that $\{a_{2n}\}$ is a decreasing sequence, and that $\{a_{2n-1}\}$ is an increasing series, and that $\lim_{n\rightarrow \infty}(a_{n+1}-a_n)=0$, and so by Cantor's intersection theorem (http://en.wikipedia.org/wiki/Cantor%27s_intersection_theorem), $\{a_n\}$

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