Prove that the sequence $x_{n}$ converges to zero

Question

We are given that $x_{n+1}=\sqrt{x_n + 1} -1$ for $n\geq 1$ and $x_1\in (-1,0)$.

Previous parts of the question have already proven that $x_n\in(-1,0)$, that $(x_n)$ is an increasing sequence, and that $(x_n)$ converges to some limit, but I'm having difficult proving that the limit is zero.

This is an introductory analysis course, so at this point they mainly want us to prove using $\epsilon$/N proofs for sequences. I've mostly tried to prove by showing that a nonzero limit gives a contradiction, but haven't had any success.

Answer 1

Since you proved that the sequence converges to a limit $\ell$, just let $n$ go to $+\infty$ in the relation that defines the sequence. You obtain $$\ell=\sqrt{\ell+1}-1$$ and from here it is not difficult to obtain that the limit is $0$.

Answer 2

If you showed that $x_n$ is increasing , contained in $(-1,0)$, and converges to some number $L$, then you know that:

(i) $x_n\leq L$

(ii) $-1\leq L\leq 0$

(iii) By uniqueness of the limit you know that $L=\sqrt{L+1}-1$, which is equivalent to $L(L+1)=0$.

Answer 3

First put $$x_{n-1} = -1 + \sqrt{x_{n-2} + 1}$$ into $$x_n = -1 + \sqrt{x_{n-1} + 1}$$ You'll notice a general form:

$$x_n = -1 + (x_{n-r} + 1)^{2^{-r}}$$

Then put r = n-1, and take the limit of both sides with n tending to infinity. On the right hand side, you have $$-1 + \lim_{n\to \infty} (x_1 + 1)^{1\over 2^{n-1}}$$ The power approaches 0, the limit approaches 1 and the final answer approaches 0.