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

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 Answer

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 $$x$$.