Do these two sequences converge to same limit? $(x_n)_{n\geqslant1}$ 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?
 A: 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
$$\frac{y_n}{y_{n+1}} =\frac{1}{2}\left(1+\frac{y_n}{x_n}\right)$$
A: $$
\begin{align}
x_{n+1}-y_{n+1}
&=\frac12\left(x_n+y_n\right)-\frac{2x_ny_n}{x_n+y_n}\\
&=\frac12\frac{(x_n-y_n)^2}{x_n+y_n}\tag{1}
\end{align}
$$
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
$$
\lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n=\sqrt{x_1y_1}\tag{7}
$$
A: 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 $l=m$.
