# Showing that a recurrence relation is increasing

Suppose $$(x_n)_{n\in\mathbb{N}}\in\mathbb{R}^{\mathbb{N}}$$ satisfies $$x_{n+1}=\sqrt{x_n+1}-1$$ and that for all $$n\in\mathbb{N}$$, $$x_n\in(-1,0).$$

How would I show that the sequence described by this recurrence relation is increasing?

• You have that $(x_{n+1}+1)^2=x_n+1$, but $0< x_{n+1}+1 <1$, so $$x_{n+1}+1>(x_{n+1}+1)^2=x_n+1,$$ then $x_{n+1}>x_n$. – guchihe Mar 6 at 1:52

Letting $$y_n = -x_n$$, then $$x_{n+1} =\sqrt{x_n+1}-1$$ becomes $$-y_{n+1} =\sqrt{-y_n+1}-1$$ or $$y_{n+1} =1-\sqrt{1-y_n}$$.
Therefore $$1-y_{n+1} =\sqrt{1-y_n}$$.
Letting $$z_n = 1-y_n$$, this is $$z_{n+1} =\sqrt{z_n}$$.
If $$0 < z_n < 1$$, then $$0 < z_{n+1} < 1$$ and $$z_{n+1} \gt z_n$$ so $$1-y_{n+1} \gt 1-y_n$$ so $$y_{n+1} \lt y_n$$ so $$x_{n+1} \gt x_n$$.