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Given positive $a$ and $b$ with $a < b$, define two sequences recursively with $x_0 = a$, $y _0 = b$, $y_{n+1} = G(x_n, y_n)$, $x_{n+1} = H(x_n, y_{n+1})$. Prove that both sequences converge. $G$ and $H$ are the geometric mean and harmonic mean, respectively.

Probably need to start off with proving both sequences are monotone and have an upper bound, but have no idea where to take the proof from that point. Any hints would be greatly appreciated.

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Claim: $x_n\le y_n$

For $n=0$ this is clear. Assume that it holds for $n$. Then, by definition, $y_{n+1} = \sqrt{\frac{y_n}{x_n}}x_n\ge x_n$. Hence, $$ x_{n+1} = \frac 2 {1+\frac{y_{n+1}}{x_n}} y_{n+1}\,\le\,y_{n+1}. $$ Done. Hence, $y_{n+1} = \sqrt{\frac{x_n}{y_n}}y_n\le y_n$ for all $n$. Moreover, we make use of $y_{n+1}\ge x_n$ from above to see that $x_{n+1} = \frac{2}{1+\frac{x_n}{y_{n+1}}}x_n\ge x_n$.

Hence, $x_n$ increases and is bounded by $y_n\le y_0 = b$ and $y_n$ decreases and is bounded below by $x_n\ge x_0=a$. So, $x_n\to\alpha$ and $y_n\to\beta$. But $$ \beta = \lim_{n\to\infty}y_{n+1} = \lim_{n\to\infty}\sqrt{x_ny_n} = \sqrt{\alpha\beta}, $$ which implies $\alpha = \beta$.

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You can show, by induction, that both sequences are bounded : they stay between $a$ and $b$. Hence, both sequences converge, up to a subsequence. You can then try to prove that there is only one possible limit to each of these sequences. Haven't tried it on paper though.

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Without loss of generality, consider the initial iteration.

$y_1=\sqrt {ab}$ and $x_1= \frac {2a\sqrt {ab}}{a+\sqrt {ab}}$.

Then $y_1-x_1=\sqrt {ab}(\frac{\sqrt {b}-\sqrt {a}}{\sqrt {b}+\sqrt {a}})=\frac{\sqrt {ab}}{({\sqrt {b}+\sqrt {a}})^2}(b-a)<\frac{1}{2}(b-a).$

Thus the difference between the $x$ and $y$ terms at least halves at each iteration and therefore tends to $0$.

Note that $b\ge y_1 \ge a$ and that $ y_1\ge x_1 \ge a$. Therefore the $x$ sequence is m.i. to the common limit and the $y$ sequence is m.d. to the common limit.

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  • $\begingroup$ This is actually used to prove in the second part of the problem that the sequences converge to the same number $\endgroup$ – carbonv2 Oct 8 '19 at 21:30
  • $\begingroup$ It can be used to prove both parts together as in the revised version given above. $\endgroup$ – S. Dolan Oct 8 '19 at 21:50

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