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Let $x_1>2$ and $x_{n+1} := 1 + \sqrt{x_n - 1}$ for all $n \in \mathbb{N}$. Show that $\{x_n\}$ is decreasing and bounded below, and find its limit.

I showed that it's bounded below ($x_n>2$ for all $n$) using induction. But then I don't know how to do the rest. Any help would be much appreciated.

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  • $\begingroup$ Rewrite this as a recursion over $y_n=x_n-1$. $\endgroup$ – Did Oct 13 '14 at 20:54
  • $\begingroup$ Is $x_1>1$????? $\endgroup$ – AsdrubalBeltran Oct 13 '14 at 20:57
  • $\begingroup$ Sorry, it's $x_1 >2$ $\endgroup$ – Shad0wNinja Oct 13 '14 at 21:22
  • $\begingroup$ Whenever they say "show the the sequences converges and find its limit" they really mean "find its limit and then show that the sequence converges". $\endgroup$ – DanielV Oct 13 '14 at 23:07
  • $\begingroup$ Possible duplicate of Let $x_1 \ge 2, x_{n+1}=1+\sqrt{x_n-1}$. Find the limit, if it is convergent. $\endgroup$ – Arnaud D. Apr 18 at 12:12
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Any recursively defined sequence where the recursive relation (in this case $1+\sqrt{x_n - 1}$) is continuous can only ever converge to $x$ which are stationary points, i.e. points so that if you enter $x$ into the relation, you get $x$ back.

This means that if your sequence converges to some $x$, then it must be the case that $$x=1+\sqrt{x-1}$$

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Clearly $$ x_{n+1}-1=\sqrt{x_n-1}=\sqrt[4]{x_{n-1}-1}=\cdots=(x_1-1)^{2^{-n}}\to 1, $$ and hence $$ x_n\to 2. $$

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If you already prove that $x_n> 2$, then $$\dfrac{x_{n+1}-1}{x_n -1} = \dfrac{1}{\sqrt{x_n -1}} < 1 $$ $$\implies x_{n+1} -1 < x_n -1$$ So $x_n$ is decreasing and bounded below. The limit point is already in the other answers.

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