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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.
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 $$x=1+\sqrt{x-1}$$
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 $$ x_n\to 2. $$
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.
