real analysis - How does ($x_n$) converge for $x_1=1$, $x_{n+1}=frac{1}{x_n+3}$ for $n=1,2,ldots$?

Show that the sequence ($x_n$) defined by

$$x_1=1\quad \text{and}\quad x_{n+1}=\frac{1}{x_n+3} \quad (n=1,2,\ldots)$$ converges and determine its limit ?

I try to show ($x_n$) is a Cauchy sequence or ($x_n$) is decreasing (or increasing) and bounded sequence but I fail every step of all.

Answer

Hint: For $x,y \geq 0$ we have $\left\vert\frac{1}{x+3} - \frac{1}{y+3}\right\vert = \left\vert\frac{y-x}{(x+3)(y+3)}\right\vert \leq \frac{1}{9}\vert x-y\vert$.