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Problem $\mathbf{4}$ If $x_{n+1}=\frac{x_n+y_n}2$, $y_{n+1}=\frac{2x_ny_n}{x_n+y_n}$ for $n=1,2,\ldots,n$, prove that both $\{x_n\}$ and $\{y_n\}$ are convergent, and $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}y_n$.

Solution: Note that $$4x_ny_n\le(x_n+y_n)^2$$ Hence, $$\frac{y_{n+1}}{x_{n+1}}=\frac{4x_ny_n}{(x_n+y_n)^2}\le 1$$ That is, $\color{crimson}{x_n\le y_n}$. Also, $$x_{n+1}-x_n=\frac{x_n+y_n}2-x_n=\frac{y_n-x_n}2\ge 0$$ implying that $\{x_n\}$ is monotonically increasing. Similarly, $$\frac{y_{n+1}}{y_n}=\frac{2x_n}{x_n+y_n}\le\frac{2x_n}{2x_n}=1$$ implying that $\{y_n\}$ is monotonically decreasing. Now we have $$x_1\le x_2\le\cdot\le x_n\le\cdot\le y_n\le\cdots\le y_2\le y_1$$ By the monotone convergence theorem, it follows immediately that both $\{x_n\}$ and $\{y_n\}$ are convergent. Let $$\lim_{n\to\infty}x_n=x,\lim_{n\to\infty}y_n=y$$ Taking limits on both sides of $x_{n+1}$ will yield that $$x=\frac{x+y}2$$ which implies that $\lim\limits_{n\to\infty}x_n=x=y=\lim\limits_{n\to\infty}y_n$

(Original image here.)

I have a question about how I can know that $x_n\le y_n$ (in red above).

I know that $y_{n+1} < x_{n+1}$, but how do I know $x_n\le y_n$?

Also, what is meant by ‘Taking limits on both sides of $x_{n+1}$’?

(Questioned parts highlighted in original image.)

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  • $\begingroup$ What are $\;x_1,\,y_1\;$ ? Clearly the red part depends on this. Is there any more information you're withholding? $\endgroup$ – DonAntonio Oct 21 '16 at 15:13
  • $\begingroup$ This seems to be arithmetic-harmonic mean. $\endgroup$ – Martin Sleziak Oct 21 '16 at 15:24
  • $\begingroup$ @DonAntonio: We don’t need $x_1$ and $y_1$ for the theorem; the red inequality is backwards and needs to be restricted to $n\ge 2$, that’s all. $\endgroup$ – Brian M. Scott Oct 21 '16 at 15:24
  • $\begingroup$ @BrianM.Scott I'm not sure. What is shown there is $\;\cfrac{y_{n+1}}{x_{n+1}}\le 1\;$ , and from here they get (in red) $\; x_n\le y_n\;$ ...as if $\;x_n\;$ is negative. $\endgroup$ – DonAntonio Oct 21 '16 at 15:36
  • $\begingroup$ @DonAntonio: I think that that’s simply a mistake. However, you raise a good point anyway: the argument that I think was intended appears to rely on having $x_{n+1}>0$, which in turn does mean that they probably intended to require that $x_1\ge 0$ and $y_1>0$. $\endgroup$ – Brian M. Scott Oct 21 '16 at 15:45
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The solution is simply wrong. The first step actually shows that $y_n\le x_n$ for $n\ge 2$. (Note that we do have to rule out $n=1$, since it’s possible that the initial values $x_1$ and $y_1$ don’t satisfy this inequality.)

Added: In fact to draw this conclusion we need to know that $x_{n+1}>0$, and at other points in the argument there appears to be an implicit assumption that $x_n$ and $y_n$ are positive. This will be the case if we add the assumption that $x_1\ge 0$ and $y_1>0$. A similar argument works if $x_n$ and $y_n$ are always both negative.

The second step should then be

$$x_{n+1}-x_n=\frac{x_n+y_n}2-x_n=\frac{y_n-x_n}2\le 0\;,$$

so that $\{x_n\}$ is monotonically decreasing. The next step then becomes

$$\frac{y_{n+1}}{y_n}=\frac{2x_n}{x_n+y_n}\ge\frac{2x_n}{2x_n}=1\;,$$

so that $\{y_n\}$ is monotonically increasing, and we have

$$y_1\le y_2\le\ldots y_n\le\ldots\le x_n\le\ldots x_2\le x_1\;.$$

It still follows from the monotone convergence theorem that both sequences are convergent, and we can let their limits be $x$ and $y$, as in the given solution. The final step is also correct, but I’ll expand on it to make clearer just what is being done. The point is that

$$\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}x_n\;,$$

so that

$$x=\lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}\frac{x_n+y_n}2=\frac{x+y}2\;.$$

Thus, $2x=x+y$, and $x=y$.

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This sequence is known under the name "arithmetic-harmonic mean". See (Proof the the Arithmetic-Harmonic Mean is expressible as the Geometric Mean) where it is proven that these sequences have a common limit which is the geometrical mean of the initial terms: $x=y=\sqrt{x_1y_1}$.

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