What is a direct method for finding the supremum of the following inductively defined sequence?

Question

The sequence is $z_1=1, z_{n+1}=\sqrt{2z_n}$. If $\epsilon>0$ is given, we must find $m\in\mathbb{N}$ such that $2-\epsilon<2^{1-2^{1-m}}$. But I'm not sure how to find a suitable $m$. I've tried taking a natural log, but that didn't appear to simplify things. I've attached an image of the work I've done so far.

Answer 1

To actually find a suitable $m$, require $2 - \epsilon < 2^{1-2^{1-m}}$, so $1 - \frac{1}{2}\epsilon <2^{-2^{1-m}}=4^{-2^{-m}}$. Now you can take $\log$ of both sides twice: $-\ln(1 - \frac{1}{2}\epsilon) >2^{-m}\ln4$, then $-\ln\Big(-\dfrac{\ln(1 - \frac{1}{2}\epsilon)}{\ln4}\Big) <m\ln2$. If you want to tidy this up, $-\ln(1 - \frac{1}{2}\epsilon)>\frac{1}{2}\epsilon$, so the requirement is $-\ln\Big(\dfrac{\epsilon}{2\ln4}\Big) <m\ln2$. Also $-\ln\Big(\dfrac{\epsilon}{2\ln4}\Big) <-\ln\Big(\dfrac{\epsilon}{4}\Big)$, so $m>-\log_2\Big(\dfrac{\epsilon}{4}\Big)=2-\log_2 \epsilon$ will do.

A direct method for getting this is to realise that, because $2 < e$, you have $2^{-x} > 1-x$ (for $x>0$), so you could solve $2^{- \frac{1}{2}\epsilon} <2^{-2^{1-m}}$.