Proof that rational sequence converges to irrational number Let $a>0$ be a real number and consider the sequence $x_{n+1}=(x_n^2+a)/2x_n$.
I have already shown that this sequence is monotonic decreasing and thus convergent, now I have to show that $(\lim x_n)^2 = a$ and thus exhibit the existence of a positive square root of $a$. (because we took $x_1 > 0$
 A: You've done the hardest part. Let $\displaystyle\lim_{n\rightarrow\infty}x_n=x$. Rearranging the recursion formula gives
$$2\cdot x_n\cdot x_{n+1}=x_n^2+a$$
Taking limits and noting that $\displaystyle \lim_{n\rightarrow\infty}x_n=\lim_{n\rightarrow\infty}x_{n+1}=x$, we have
$$
2x^2=x^2+a
$$
so $x^2=a$.
A: If $x_n\to x$, then 
$$
x_{n+1}=\frac{x_n^2+a}{2x_n} \to a
$$
as well. But
$$
\frac{x_n^2+a}{2x_n} \to \frac{x^2+a}{2x}.
$$
Thus the limit $x$ satisfies the equation
$$
x=\frac{x^2+a}{2x} \quad\Longleftrightarrow\quad 2x^2=x^2+a
\quad\Longleftrightarrow\quad x^2=a \quad\Longleftrightarrow\quad x=\pm \sqrt{a}.
$$
But as the sequence has terms terms it can not converge to a negative number. Thus $x_n\to\sqrt{a}$.
A: The recursive definition can be solved exactly. Write a^2 instead of a and consider the sequence $(x(n)+a))/(x(n)-a)$.
Then $(x(n)+a)/(x(n)-a) =
= (.5*x(n-1)+.5*a^2/x(n-1)+a)/(.5*x(n-1)+.5*a^2/x(n-1)-a) =
= (x(n-1)^2+2*a*x(n-1)+a^2)/(x(n-1)^2-2*a*x(n-1)+a^2) =
= ((x(n-1)+a)/(x(n-1)-a))^2 =$
Thus $(x(n)+a)/(x(n)-a) = ((x(0)+a)/(x(0)-a))^{(2^n)}$.
This proves (very) rapid convergence to +a whenever $|x(0)-a)|<|x(0)+a)|$ and to -a whenever$|x(0)+a)|<|x(0)-a)|$. This holds even when a is complex and complex sequences are considered. You could work out, for instance, the case of a purely imaginary root and real starter $x(0)$...
PS: This works only in this (Babylonian) case.  
