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If $a_1=1/2$ and $a_{n+1} = a_n^2$, prove that this recursive sequence is convergent.

I know I need to show that it is bounded and monotone decreasing, but I'm not sure how to go about doing that.

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  • $\begingroup$ You start with $a_i$, which is positive, and less than $1$. You multiply it with itself to get $a_{i+1}$. The result becomes smaller, but still positive, wouldn't you say? $\endgroup$
    – Arthur
    Commented Apr 15, 2016 at 15:29

4 Answers 4

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The sequence is obviously bounded below by $0$, because $a_{n+1}=a_n^2\ge0$ and $a_1\ge0$.

Next, prove that $a_{n+1}\le a_n$, that is, $a_n^2\le a_n$. Since these are nonnegative numbers, this reduces to showing that $a_n\le 1$, for all $n$.

The base step is obvious. For the induction step use that $0\le x\le 1$ implies $x^2\le1$.

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Let $f(x)=x^2$, you have $a_{n+1}=f(a_n)$. You notice that $f([0,1])=[0,1]$ and since $a_0 \in [0,1]$, $\forall n, a_n \in [0,1]$. Also on $[0,1]$, $f(x) \le x$, so $a_n$ is decreasing and has a lower bound ($0$). So the sequence is convergent.

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$a_{n+1}=a_n$ with $a_1=1/2$.

Then $a_2=\frac{1}{2}^2$

$a_3=((\frac{1}{2})^2)^2=\frac{1}{2}^4$

$a_4=(a_3)^2=(\frac{1}{2}^4)^2=\frac{1}{2}^8$

and in general: $a_n=(\frac{1}{2})^{2^{n-1}} \leq (\frac{1}{2})^n$

It should be much easier to prove that $\{\frac{1}{2}^n\}$ converges to $0$.

edit The sequence is bounded below by $0$ and $a_{n+1}<a_n$ so it is strictly monotone as well.

To show it's bounded below by zero, just note that $2^{anything}>0$.

To show that it's monotone, for any $0<x<1$, you want to show that $x^2<x$.

Well, $x<1$, meaning that $x^2=x\cdot x<1 \cdot x=x$.

Since $0<a_1<1$, the result follows immediately.

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Well prove with Inductive reasoning that for all $n \in \mathbb{N}$, $0 \leq a_{n} \leq 0.5$:

for $n=1$, $0 \leq a_{1}=0.5 \leq 0.5$

we suppose that the inequality is true for $n=k$: $0 \leq a_{k} \leq 0.5$

so we can understand that $0 \leq a_{k+1}=a_{k}^2 \leq0.25\leq 0.5$

that's mean that the inequality is true for $n=k+1$

We proved that the sequence $(a_{n})$ is bounded.

Now we can understand that for all $n \in \mathbb{N}$, $\frac{a_{n+1}}{a_{n}}=a_{n}\leq 1$ By that we can understand that the sequence $(a_{n})$ is monotonically decreasing. If sequence is bounded and monotonic, it convergent.

*Sorry for my English.

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