$a_1>0$
$a_{n+1} = {1\over2}(a_n + {1 \over a_n})$
assume $\lim a_n = A \in \mathbb R - \{0\}$
Can anybody help me solve this? We should first show the limit exists. That should be easiest by showing that the sequence is monotonous but in this case it doesn't really look monotonous and I am stuck.
Thanks.
On
Knowing $\lim_{n\to\infty} a_n = 1$ from $x=\frac12\left(x+\frac1x\right)$, \begin{align} &b_n = a_n - 1\\ &b_{n+1}=\frac12\left(b_n+1+\frac1{b_n+1}\right)-1=\frac12\frac{{b_n}^2+2b_n+2-2b_n-2}{b_n+1}=\frac12\frac{b_n^2}{b_n+1}\\ &c_{n+1}=\frac1{b_{n+1}}=2c_n^2+2c_n\\ &c_{n+1}+\frac12=2\left(c_n+\frac12\right)^2=2^2\left(c_{n-1}+\frac12\right)^4=\cdots=2^{n+1}\left(c_0+\frac12\right)^{2^{n}}\\ &c_n=2^n\left(c_0+\frac12\right)^{2^{n-1}}-\frac12=2^n\left(\frac{1}{a_0-1}+\frac12\right)^{2^{n-1}}-\frac12\\ &\therefore a_n=b_n+1=\frac1{c_n}+1=\frac1{2^n\left(\frac{1}{a_0-1}+\frac12\right)^{2^{n-1}}-\frac12}+1\\ \end{align} We can confirm that $\lim_{n\to\infty} a_n = 1$.
From the inequality between the arithmetic and geometric mean it's easy to see that $$x+\frac{1}{x} \geq 2$$ for every $x>0$ so we can deduce that $a_n\geq 1$ for every $n \geq 2$ .
Now $a_n$ is monotonically decreasing for $n \geq 2$ because : $$a_{n+1}-a_n=\frac{1}{2} \left (\frac{1-a_n^2}{a_n} \right ) \leq 0$$
The sequence is thus lower bounded by $1$ and monotonically decreasing so it's convergent to a number , let's say $A$ .
Now let $n \to \infty$ in the recurrence to get : $$A=\frac{1}{2} \left (A+ \frac{1}{A} \right )$$ $$A=\frac{1}{A}$$ $$A^2=1$$ so the sequence must be convergent to $A=1$ .