# How to prove that the limit belongs to $(0, 1)$? [closed]

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 $$(0, 1)$$?

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

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.

• Thank you for your answer, but I can't understand why did you take L/8 for epsilon? – TOOF4CK Dec 11 '19 at 11:28
• 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 – Omnomnomnom Dec 11 '19 at 11:29
• Now I got this, thank you! – TOOF4CK Dec 11 '19 at 11:31

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.

• It bothers me that the wolf-ears don't lineup with the unicorn-ears. Otherwise, great answer. – Omnomnomnom Dec 11 '19 at 10:38
• @Omnomnomnom: Is it better now? – Martin R Dec 11 '19 at 10:42
• I definitely think so $\ddot\smile$ – Omnomnomnom Dec 11 '19 at 10:46
• Sorry for my silly question but why An = (An-1)/2 is not possible for this example? – TOOF4CK Dec 11 '19 at 13:41
• @TOOF4CK: I have updated to answer to clarify that. – Martin R Dec 11 '19 at 14:26

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 $$\lim_{n\to\infty} a_n = 0$$
• 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. – Paul Sinclair Dec 11 '19 at 18:31