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?

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    $\begingroup$ 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 $x_{n+1}>x_n$. $\endgroup$ – 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 $.


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