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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.

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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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  • $\begingroup$ equations did not show up properly. please check! It is OK now. Thanks! $\endgroup$ – mike Sep 11 '14 at 5:43
  • $\begingroup$ @mike: I'm not sure what the problem was. I didn't make any edits. Maybe there was an issue with your browser. $\endgroup$ – Bungo Sep 11 '14 at 5:49
  • $\begingroup$ you are right. it is OK now. $\endgroup$ – mike Sep 11 '14 at 6:01
  • $\begingroup$ Wow .. nice and intuitive solution :-) $\endgroup$ – MathMan Sep 14 '14 at 5:36

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