How prove that $0 < a_{10} - \sqrt{2}< 10^{-370}$ for $a_n = \frac{1}{2}(a_{n-1} + \frac{2}{a_{n-1}})$? Let $a_1=1,$ , $a_n = \frac{1}{2}(a_{n-1} + \frac{2}{a_{n-1}})$. How prove that
$0 < a_{10} - \sqrt{2}< 10^{-370}$?
$a_n - a_{n-1} = \frac{1}{2a_{n-1}} \left(2 - a_{n-1}^2\right) < 0 \Rightarrow a_{n-1} > \sqrt{2}$ and what next?

We have
$$a_{n+1} - \sqrt2 = \dfrac12\left(a_n-\sqrt2 + \dfrac2{a_n} - \sqrt2\right) = \dfrac12\left(a_n-\sqrt2 + \dfrac{2-\sqrt2a_n}{a_n}\right) = \dfrac{(a_n-\sqrt2)^2}{2a_n}$$
Clearly, $a_n > \sqrt2$ for all $n >1$. Hence, we obtain that
$$a_{n+1} -\sqrt2 \leq \dfrac{(a_n-\sqrt2)^2}{2\sqrt2} \leq \dfrac{(a_{n-1}-\sqrt2)^4}{(2\sqrt2)^{1+2}} \leq \dfrac{(a_{n-2}-\sqrt2)^8}{(2\sqrt2)^{1+2+4}} \leq \cdots \leq \dfrac{(a_2-\sqrt2)^{2^{n-1}}}{(2\sqrt2)^{2^{n-1}-1}}$$
Hence, we have
$$a_{10} - \sqrt2 \leq \dfrac{(3/2-\sqrt2)^{2^8}}{(2\sqrt2)^{2^8-1}} < 10^{-370}$$
where the last inequality can be checked using logarithms.
