Proving convergence of a recursively defined sequence with $\sqrt{2}$

111 Views Asked by At

I'm currently working on a recursive sequence where I am meant to prove that it converges but I haven't done this stuff for years and I'd love it if someone could give me a push in the right direction.

The sequence is defined as $x_1=1$ and $x_{n+1} = (\sqrt{2})^{x_n}$

There is a hint to use induction and show that $x_n < 2$ but I'm not sure how to form an induction proof for something like this.

I figure the first few terms would be $ 1, (\sqrt{2}), {\sqrt{2}}^{\sqrt{2}}, {\sqrt{2}}^{{\sqrt{2}}^{\sqrt{2}}}$ etc. but to prove that it is $x<2$ for all of $n$ I would need to prove it true for $n+1$, correct? I'm not sure how I can write the sequence in a way that I can prove this.

Any help or tips would be much appreciated.

1

There are 1 best solutions below

0
On

To show this by induction, you need to prove that if $x_n < 2$, then $x_{n+1} < 2$. (You also need to show $x_1 < 2$, but that's rather obvious.) In other words, you need to show that $$ x_n < 2\Longrightarrow (\sqrt{2})^{x_n} < 2 $$

As a hint, remember that if $a > 1$, $a^x$ is an increasing function of $x$.