how can I prove it is a convergent sequence and how can I calculate its limit

Question

Let we have the following sequence $$x_1=1$$ $$x_{n+1}=\frac{1}{2+x_n}$$ how can I prove it is a convergent sequence and how can I calculate its limit

Answer 1

Notice that $f(x) = \frac{1}{2 + x}$ is a strict contraction on $[0,\infty)$, indeed $|f'(x)| = \frac{1}{(2 + x)^2} \le \frac14$. Then, by the Contraction mapping theorem, $x_{n + 1} = f(x_n)$ is a Cauchy sequence which converges to the unique positive fixed point of $f$: $$L = \frac{1}{2 + L} \Longrightarrow L = \sqrt2 - 1.$$

Answer 2

Let $L=\sqrt{2}-1$. Observe that $x_n>0$ for all $n$. Therefore

$$ \begin{align*} |x_{n+1}-L|&=\left|\frac{1-L(2+x_n)}{2+x_n}\right|\\ &=\left|\frac{L}{2+x_n}(L^{-1}-(2+x_n))\right|\\ &=\left(\frac{L}{|2+x_n|}\right)|L-x_n|\\ &\leq \frac{L}{2}|L-x_n|. \end{align*} $$ Since $L/2<1$, it follows that $\lim_{n\to\infty}x_n=L$.