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In a sequence of positive numbers $$a_{1}, a_{2},\dots,a_{n},\dots,$$

each of the terms $a_n$ is equal to either $\dfrac{ a_{n-1} }{ 2}$ or $\sqrt{a_{n - 1}}$.

Can this sequence have a limit belonging to the interval $(0, 1)$?

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    $\begingroup$ What is $\sqrt{a_n-1}$ for $a_n<1$?. Did you mean $\sqrt{a_{n-1}}$? $\endgroup$ – Maximilian Janisch Dec 11 '19 at 9:56
  • $\begingroup$ what does "Can this sequence have a limit" mean? does it mean $\lim_{n\to\infty}a_n$? $\endgroup$ – John Dec 11 '19 at 9:59
  • $\begingroup$ @John in other words: the question is whether there exist a sequence satisfying these criteria such that $$ 0 < \lim_{n \to \infty} a_n < 1 $$ $\endgroup$ – Omnomnomnom Dec 11 '19 at 10:03
  • $\begingroup$ @Omnomnomnom, shouldn't we define what $a_n$ is beforehand? $\endgroup$ – John Dec 11 '19 at 10:06
  • $\begingroup$ @John no: the question is about all sequences satisfying these criteria. $\endgroup$ – Omnomnomnom Dec 11 '19 at 10:07
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I assume that you meant to say that for each $n$, we have either $a_n = \frac 12 a_{n-1}$ or $a_n = \sqrt{a_{n-1}}$.

The answer to your question (as I understand it) is no. Suppose for the purpose of contradiction that $\lim_{n \to \infty} a_n = L$ with $0 < L < 1$. First, note that there must be infinitely many $n$ such that $a_{n} = \frac 12 a_{n-1}$. Otherwise, the sequence would converge to a fixed point of the function $f(x) = \sqrt{x}$, namely $0$ or $1$.

Let $N$ be such that $n > N$ implies that $|a_n - L| < L/8$. Select an $n>N$ such that $a_{n+1} = \frac 12 a_n$. We note that $$ |a_{n+1} - L| \geq |a_{n+1} - a_n| - |a_n - L| = \frac 12 a_n - |a_n - L| \\ > \frac 12 (L - L/8) - L/8 = \frac 5{16}L > L/8. $$ However, our definition of $N$ implies that since $n+1 > N$, we have $|a_{n+1} - L| < L/8$. So, we have a contradiction.

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  • $\begingroup$ Thank you for your answer, but I can't understand why did you take L/8 for epsilon? $\endgroup$ – TOOF4CK Dec 11 '19 at 11:28
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    $\begingroup$ I just had to make it small enough for the last line of the inequality to work. With $L/4$ I wouldn't have been able to use the triangle inequality like I did $\endgroup$ – Omnomnomnom Dec 11 '19 at 11:29
  • $\begingroup$ Now I got this, thank you! $\endgroup$ – TOOF4CK Dec 11 '19 at 11:31
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Assume that $\lim_{n \to \infty} a_n = a$ exists, with $0 < a < 1$. Then there exists an $N$ such that for all $n \ge N$ $$ \frac 34 a < a_n < \frac 54 a \, . $$ It follows that for $n> N$ $$ \color{red}{a_n} > \frac 34 a > \frac 12 \cdot \frac 54 a \color{red}{> \frac 12 a_{n-1}} $$ and consequently $$ a_n = \sqrt{a_{n-1}} \, . $$ The limit must therefore satisfy $$ a = \sqrt a $$ which is not possible for $0 < a < 1$.

Therefore such a sequence does not exist.

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  • $\begingroup$ It bothers me that the wolf-ears don't lineup with the unicorn-ears. Otherwise, great answer. $\endgroup$ – Omnomnomnom Dec 11 '19 at 10:38
  • $\begingroup$ @Omnomnomnom: Is it better now? $\endgroup$ – Martin R Dec 11 '19 at 10:42
  • $\begingroup$ I definitely think so $\ddot\smile$ $\endgroup$ – Omnomnomnom Dec 11 '19 at 10:46
  • $\begingroup$ Sorry for my silly question but why An = (An-1)/2 is not possible for this example? $\endgroup$ – TOOF4CK Dec 11 '19 at 13:41
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    $\begingroup$ @TOOF4CK: I have updated to answer to clarify that. $\endgroup$ – Martin R Dec 11 '19 at 14:26
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The answer is: No.

Either it holds $a_n \ge 1$ for all $n\in\Bbb{N}$ then obviously $$\lim_{n\to\infty} a_n \ge 1$$ if it exists.

Or there is one $n_0 \in \Bbb{N}$ s.t. $a_n < 1$ but then $\sqrt{a_n -1}$ is not defined so for all following elements of the sequence it has to hold $$a_n = \dfrac{ a_{n-1} }{ 2}$$ but then obviously $$\lim_{n\to\infty} a_n = 0$$

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  • $\begingroup$ For any late readers, the original posting of the question had $a_n = \sqrt{a_n-1}$ as a typo, which was only corrected after Gono's posting of this answer. $\endgroup$ – Paul Sinclair Dec 11 '19 at 18:31

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