<s>[INST] Prove that $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ converges when $n \to \infty$ I want to prove that the sequence defined by $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ has a limit.
By evaluating the sequence I notice that the sequence is strictly monotonically decreasing starting from $x_2=1.5$.
It seems to suggest itself to prove that the sequenced is bounded by $1\le\big( x_n \big)_{n\ge1} \le 1.5$ and to prove that it is strictly monotonically decreasing starting at $x_2$ which would imply convergence.
How would I proceed and could one prove the existence of the limit without first evaluating the values of the sequence to see how the sequence behaves?
 [/INST] A. $x_n>0$ , $n\in \mathbb{N}$.
B.
We will show that $\{x_2, x_3, \ldots, x_n, \ldots \}$ is monotonically non-increasing.
$$
x_n-x_{n+1} = x_n - \frac{x_n}{2} - \frac{1}{x_n}  = \frac{x_n}{2} - \frac{1}{x_n} =
\frac{x^2_n-2}{2x_n}, \qquad n>1. 
$$
But $x^2_n-2>0$, when $n>1$, because
$$
x^2_n-2 = \left( \dfrac{x_{n-1}}{2}+\dfrac{1}{x_{n-1}} \right)^2 - 2 = 
\dfrac{x^2_{n-1}}{4} + 1 + \dfrac{1}{x^2_{n-1}} - 2 = 
$$
$$
\dfrac{x^2_{n-1}}{4} - 1 + \dfrac{1}{x^2_{n-1}}  =
 \left( \dfrac{x_{n-1}}{2}-\dfrac{1}{x_{n-1}} \right)^2  \geqslant 0.
$$
So, 
$$x_n-x_{n+1} \geqslant 0.$$
Proved.
C. Since the sequense has the limit (denote $X$) $-$ as bounded and non-increasing sequence, then
$$
X = \frac{X}{2}+\frac{1}{X};
$$
$$
X^2 = \frac{X^2}{2}+1;
$$
$$
\frac{X^2}{2}=1;
$$
$$
X = \sqrt{2}.
$$
</s>