<s>[INST] If $a_{1}=1$ and $b_{1}=2$ show that $b_{5}-a_{5}<1/2^{45}$. Need help interpreting solution. If $a_{1}=1$ and $b_{1}=2$ show that $b_{5}-a_{5}<1/2^{45}$.
$a_{n+1}=\sqrt{a_{n}b_{n}}$  , and $b_{n+1}=\frac{1}{2}(a_{n}+b_{n})$.
I've managed to work out that $b_{n+1}-a_{n+1}<\frac{1}{8}(b_{n}-a_{n})^{2}$. It was the solution in the back of the book. I do not however understand how this relates to the inequality which is to be shown.
 [/INST] From
$$
b_{n+1}-a_{n+1}<\frac{1}{8}(b_{n}-a_{n})^{2},\quad n\ge1,
$$ one gets, with $n=1,2,3,4$,
$$
\begin{align}
b_{5}-a_{5}&<\frac{1}{8}(b_{4}-a_{4})^{2}
\\\\&<\frac{1}{8^3}(b_{3}-a_{3})^{2}
\\\\&<\frac{1}{8^7}(b_{2}-a_{2})^{2}
\\\\&<\frac{1}{8^{15}}(b_{1}-a_{1})^{2}.
\end{align}
$$
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