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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.

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    $\begingroup$ There's two ways I see - first, you should be able to modify your proof of the monotonicity to show that if $x_n \in (-1,0),$ then $x_{n+1}$ is strictly bigger. Use this to show that anything that isn't $0$ cannot be a limit. Alternatively, since you;ve shown that limits exist, you can simply take this limit in the recurrence relation you have. By the continuity of $\sqrt{\cdot},$ you then must have $L = \sqrt{L + 1} - 1,$ where $L$ is the limit in question. $\endgroup$ – stochasticboy321 Mar 4 '19 at 17:16
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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.

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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 $0$.

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  • $\begingroup$ This is my inclination as well, but I don't think this will be permitted as the class still takes a very elementary methodology to proving limits. We haven't proved anything this way in previous homeworks or classes, so I think they want us to continue using epsilon/N based proofs. $\endgroup$ – Kelly Marshall Mar 4 '19 at 17:15
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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 $L(L+1)=0$.

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  • $\begingroup$ Doesn't that imply that the limit could also be -1? $\endgroup$ – Kelly Marshall Mar 4 '19 at 17:16
  • $\begingroup$ If $x_1>-1$ and $x_n$ is increasing, then no. $\endgroup$ – Keen-ameteur Mar 4 '19 at 17:18
  • $\begingroup$ Okay, could expound on how we are able to rigorously prove (iii) $\endgroup$ – Kelly Marshall Mar 4 '19 at 17:22
  • $\begingroup$ $x_n$ and $x_{n+1}$ both must tend to $L$. Also one can show that $\sqrt\{ \cdot \} $ behaves well with respect to limit. As stated above this gives the equation by uniqueness of the limit. $\endgroup$ – Keen-ameteur Mar 4 '19 at 17:31
  • $\begingroup$ I suppose my issue is that for the purposes of this course, it is not sufficient to say that both $x_n$ and $x_{n+1}$ tend to the limit and therefore posit that $x_n=x_{n+1}=L$, we would have to say that for some $n>N$, we have that $ |L-x_n|<\epsilon$ and similarly for $x_{n+1}$. $\endgroup$ – Kelly Marshall Mar 4 '19 at 18:31

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