<s>[INST] Prove that $\sqrt{a_n b_n}$ and $\frac{1}{2}(a_n+b_n)$ have the same limit I'm trying to solve the following problem
prove $\sqrt{a_n b_n}$ and $\frac{1}{2}(a_n+b_n)$ have same limit. In this post https://math.stackexchange.com/a/267499, I do not understand the following steps:
$$\frac {1}{2} (\sqrt{b_n} - \sqrt{a_n} ) ( \sqrt{b_n} + \sqrt{a_n} ) = \frac{1}{2}(b_n - a_n) \leq \frac {1}{2^n} (b_1-a_1)$$
Where did the ${2^n}$ come from?
Thank you for your response.
 [/INST] Look at this chain of inequalities in the post you cited:
$$b_{n+1} - a_{n+1} = \frac {1}{2} (\sqrt{b_{n}} -\sqrt{ a_{n}})^2 \leq \frac {1}{2} (\sqrt{b_n} - \sqrt{a_n} ) ( \sqrt{b_n} + \sqrt{a_n} ) = \frac {1}{2} ( b_n - a_n)$$
So for every $n$, we have
$$b_{n+1} - a_{n+1} \leq \frac{1}{2}(b_n - a_n)$$
Therefore,
$$b_2 - a_2 \leq \frac{1}{2}(b_1 - a_1)$$
$$b_3 - a_3 \leq \frac{1}{2}(b_2 - a_2) \leq \frac{1}{2}\left(\frac{1}{2}(b_1 - a_1)\right) = \frac{1}{2^2}(b_1 - a_1)$$
It's an easy induction argument (or more informally, continuing these steps $n$ times) to conclude that
$$b_{n+1} -a_{n+1} \leq \frac{1}{2^n}(b_1 - a_1)$$
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