Prove that a sequence converges to $\sqrt{2}$ Given $x_1 = 2$ and $x_{n+1} = \dfrac{1}{2} x_n + \dfrac{1}{x_n}$, prove that $x_n \to \sqrt{2}$.
I thought I could use monotone convergence but I have a hard time proving the monotonicity of the sequence.
 A: As the sequence is positive we have that:
$$x_{n+1} \ge \sqrt{2} \iff \frac{x_n^2 + 2}{2x_n} \ge \sqrt{2} \iff (x_n - \sqrt{2})^2 \ge 0$$
Using this we have that
$$x_{n+1} - x_{n} = \frac{2-x_n^2}{2x_n} \le 0$$
Therefore the sequence is bounded from below and it's decreasing, hence it converges.
Now we can take $\lim_{n \to \infty} x_n = \lim_{n \to \infty} x_{n+1} = L$ and substitute into the equation and we'll get:
$$L = \frac{L^2 + 2}{2L} \implies L^2 = 2 \implies L=\sqrt{2}$$
A: Let's write this $x_{n+1}=\dfrac12\left(x_n+\dfrac2{x_n}\right)$. The sequence $x_n$ has actually a closed form.
First, notice that $x_n>\sqrt{2}$ for all $n$, by induction: if $x_n>\sqrt{2}$, then
$$x_{n+1}-\sqrt2=\dfrac12\left(x_n-2\sqrt{2}+\dfrac2{x_n}\right)=\frac12\left(\sqrt{x_n}-\dfrac{\sqrt2}{\sqrt{x_n}}\right)^2\geq0$$
And it can be equal to $0$ only if $x_n=\sqrt{2}$, which is wrong by hypothesis. Since $x_1>\sqrt{2}$, it's then true for all $n$.
Now, a little bit of hyperbolic trigonometry: $\coth x>1$ for $x>0$, and it's a decreasing function, from $+\infty$ (for $x\to0^+$) to $1$ (for $x\to+\infty$), hence it's a bijection from $]0,+\infty[$ to $]1,+\infty[$.
You can then find, for all $n$, a unique $\mu_n>0$ such that $x_n=\sqrt{2}\coth \mu_n$. Then simplify:
$$x_{n+1}=\dfrac12\left(\sqrt{2}\coth \mu_n+\dfrac2{\sqrt{2}\coth \mu_n}\right)=\frac{\sqrt{2}}2\left(\coth \mu_n+\tanh \mu_n\right)=\sqrt{2}\coth(2\mu_n)$$
Hence $\mu_{n+1}=2\mu_n$, and
$$x_n=\sqrt{2}\coth\left(2^{n-1}\mathrm{arccoth} (x_1)\right)$$
Since $\mathrm{arccoth} (x_1)>0$, you have $2^{n-1}\mathrm{arccoth} (x_1)\to+\infty$ as $n\to\infty$, hence
$$\lim_{n\to\infty} x_n=\sqrt{2}$$
