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$(x_n)_{n\geqslant1}$ and $(y_n)_{n\geqslant1}$ are two real sequences such that $x_1 > 0$, $y_1 > 0 $, $x_{n+1} = \frac{1}{2}(x_n + y_n)$ and $\dfrac{2}{y_{n+1}} = \dfrac{1}{x_n} + \dfrac{1}{y_n}$.

Do these two sequences converge to the same limit? If yes, what is their limit?

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3 Answers 3

if we divide 1st one by x_n

$$\frac{x_{n+1}}{x_n}=\frac{1}{2}\left(1+\frac{y_n}{x_n}\right)$$

then 2nd one multiply by y_n

$$\frac{y_n}{y_{n+1}} =\frac{1}{2}\left(1+\frac{y_n}{x_n}\right)$$

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it seems like they are different not converging to the same limit –  Jonas12 Nov 13 '12 at 9:41
    
There is a mistake in second equation –  no identity Nov 13 '12 at 9:59
    
@Norbert: I think that I fixed it. –  robjohn Nov 13 '12 at 10:06
    
It appears that $x_{n+1}y_{n+1}=x_ny_n$, so the product of the limits would be $x_1y_1$. –  robjohn Nov 13 '12 at 10:07

$$ \begin{align} x_{n+1}-y_{n+1} &=\frac12\left(x_n+y_n\right)-\frac{2x_ny_n}{x_n+y_n}\\ &=\frac12\frac{(x_n-y_n)^2}{x_n+y_n}\tag{1} \end{align} $$ If $x_n,y_n>0$, $\left|\frac{x_n-y_n}{x_n+y_n}\right|<1$, so $\frac12\frac{(x_n-y_n)^2}{x_n+y_n}<\frac12|x_n-y_n|$. Therefore, $$ 0\le x_{n+1}-y_{n+1}\le\frac12|x_n-y_n|\tag{2} $$ Thus, $$ |x_n-y_n|\le\frac1{2^{n-1}}|x_1-y_1|\tag{3} $$ Therefore, $\displaystyle\lim_{n\to\infty}x_n-y_n=0$. For $n>1$, $(1)$ guarantees that $x_{n}\ge y_n$. Furthermore, since $x_{n+1}$ and $y_{n+1}$ are means of $x_n$ and $y_n$, we have $$ y_n\le y_{n+1}\le x_{n+1}\le x_n\tag{4} $$ Therefore, $x_n$ is monotonically decreasing and $y_n$ is monotonically decreasing, and both are bounded. Thus, $$ \lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n\tag{5} $$ Furthermore, as I noted in a comment to Jonas Kgomo's answer, $$ x_{n+1}y_{n+1}=\frac12(x_n+y_n)\frac{2x_ny_n}{x_n+y_n}=x_ny_n\tag{6} $$ This easily leads to the conclusion that $$ \lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n=\sqrt{x_1y_1}\tag{7} $$

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This is a simplified version of robjohn's answer.

By the AM-Hm inequality we have $y_n \leq x_n$. Moreover

$$x_n = \frac{x_{n-1}+y_{n-1}}{2} \leq \frac{ x_{n-1}+x_{n-1}}{2}=x_{n-1}$$ and

$$y_n=\frac{2}{\frac{1}{x_{n-1}}+\frac{1}{y_{n-1}}}\geq\frac{2}{\frac{1}{y_{n-1}}+\frac{1}{y_{n-1}}}=y_{n-1}$$

This shows that $y_1 \leq y_2 \leq ..\leq y_n \leq ..\leq x_n \leq x_{n-1} \leq ... \leq x_1$$

Hence both sequences are monotonic and bounded thus convergent. Let

$$\lim x_n=l \,;\, \lim y_n =m \,.$$

Then taking the limit in $x_{n+1}=\frac{x_n+y_n}{2}$ we get $l=m$.

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In the first two inequalities, we need $n>2$ since it is possible that $x_1<y_1$. –  robjohn Nov 13 '12 at 16:21
    
It is true, that $(2)$ and $(3)$ in my answer are mostly unnecessary and $(1)$ simply justifies the AM-HM inequality. However, I do like the result in $(7)$ saying that $$ \lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n=\sqrt{x_1y_1} $$ –  robjohn Nov 13 '12 at 16:52
    
@robjohn Yes, I really like that part.. I realized while I was typing that I kinda re-say what you were saying, tried to give you credit for that; and to be honest I didn't see how to get the limit :) –  N. S. Nov 13 '12 at 19:34

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