Showing that a recursion for approximating the square root converges Let the numbers $x_n$ be defined by the recursion $$x_1=k,\quad x_{n+1}=\frac{1}{2}\bigg(x_n + \frac{k}{x_n}\bigg)\space (n\ge 2).$$ To show the numbers defined by the recursion approach $\sqrt{k}$, is it really sufficient to argue that if $x_n$ approaches $\sqrt{k}$ then so does $x_{n+1}$? Do we not have to take into account for how fast the sequence is approaching $\sqrt{k}$?
 A: WLOG, we can work with $k=1$, as the iterations can be rewritten
$$\frac{x_{n+1}}{\sqrt k}=\frac12\left(\frac{x_n}{\sqrt k}+\frac{\sqrt k}{x_n}\right),$$ or
$$y_{n+1}=\frac12\left(y_n+\frac1{y_n}\right).$$
Now
$$y_{n+1}-1=\frac{y_n^2-2y_n+1}{2y_n}=\frac{(y_n-1)^2}{2y_n}$$ and
$$\left|\frac{y_{n+1}-1}{y_n-1}\right|=\left|\frac{y_n-1}{2y_n}\right|<1$$ for $y_n>\dfrac12$.

A more powerful statement is obtained by observing that
$$\frac{y_{n+1}-1}{y_{n+1}+1}=\left(\frac{y_{n}-1}{y_{n}+1}\right)^2$$ so that by induction
$$\frac{y_{n+1}-1}{y_{n+1}+1}=\left(\frac{y_{1}-1}{y_{1}+1}\right)^{2^n}.$$
As the ratio in the RHS is always $<1$ in absolute value, its powers tend to zero, and do it very quickly. the number of exact decimals can double at each iteration. They call it a quadratic convergence speed.
A: $x_{n+1}$ is the arithmetic mean of $x_n$ and $k/x_n$, which is always at least as big as the geometric mean (AM-GM inequality), which in this case is $\sqrt{x_n\cdot k/x_n}=\sqrt k$. Therefore, $x_n$ is always an overestimate for $\sqrt{k}$. 
We can now show $x_n$ is a decreasing sequence. Since $x_n\ge \sqrt{k}$, it follows $k/x_n\le \sqrt{k}$, so that $x_{n+1}$ is the average of $x_n$ and a number less than or equal to $x_n$, so $x_{n+1}\le x_n$.
Since $x_n$ is a decreasing sequence, it must converge to some limit (instead of oscillating back and forth). Letting $x$ be that limit, and letting $n\to\infty$ in the equation $x_{n+1}=\frac12(x_n+k/x_n)$, you get $x = \frac12(x+k/x)$, which implies $x=\pm \sqrt{k}$. The negative solution can be ruled out since $x_n\ge 0$ for all $n$.
