Question: Recursively define a sequence by $x_1=1$ and $x_{n+1}=(\sqrt2)^{x_n}$. Prove that the sequence $\{x_n\}_{n=1}^{\infty}$ converges.

Attempt: To prove its convergence, I have to show the sequence is bounded and monotone.

I can prove the sequence $x_n\ge 1$ by induction.

I can prove the sequence is monotone increasing $x_n\le x_{n+1}$ by induction.

Since it is monotone increasing, I need to show the sequence is bounded above, but I don't know how to find this.

Could you give some idea? By the way, it is an assignment question.

  • $\begingroup$ Hint: Show that the property $1\leqslant x_n\leqslant2$ is hereditary. $\endgroup$ – Did Mar 23 '18 at 11:04

Note that $a^b \le a^c$ for $1\le b \le c$ and $a\ge 1$.

$|x_{2}|=(\sqrt{2})^1 \le (\sqrt{2})^2=2$.

$|x_{3}|=(\sqrt{2})^{x_2} \le (\sqrt{2})^{|x_2|} \le (\sqrt{2})^2=2$.

. . .

$|x_{n}| \le 2$.

I'm only just restating what User L KM wrote in their answer.


$x_1=1$ $$x_{n+1}=(\sqrt{2})^{x_n} \implies x_{n+1}=(\sqrt{2})^{(\sqrt{2})^{x_{n-1}}}$$ Continuing in this way we get $x_{n+1}$ as continued exponentiation upto $n=1$ For $\lim_{n \to \infty}x=(\sqrt{2})^x$
Squaring and taking log to the base 2 both sides

$$x^2=2^x\implies 2\log_2 x=x\implies x=2\log_2(2\log_2 (2\log_2......\ $$ In the innermost term there will be $\log_2 2=1$ cascading the result to the outermost term giving us $x=2$ . So the sequence converges to $x_n=2$.


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