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Let $x_1=1$ and $x_{n+1}=\sqrt{1+2/x_n}$, for $n\in\mathbb{N}$. Show that $\lim\limits_{n\rightarrow\infty} x_n$ exists and find the limit.

I have conjectured that odd-term sequence of $x_n$ is increasing and even-term sequence of $x_n$ is decreasing, but I am struggling to prove that $x_{2n+1}-x_{2n-1}>0$ and $x_{2n}-x_{2n-2}<0$, for all $n$.

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I have conjectured that odd-term sequence of $x_n$ is increasing and even-term sequence of $x_n$ is decreasing, but I am struggling to prove that $x_{2n+1}-x_{2n-1}>0$ and $x_{2n}-x_{2n-2}<0$

Hint: $\require{cancel}\;x_{n+1}^2-x_n^2 = \left(\cancel{1}+\dfrac{2}{x_n}\right)-\left(\cancel{1}+\dfrac{2}{x_{n-1}}\right)=\dfrac{-2(x_n-x_{n-1})}{x_nx_{n-1}}\,$, therefore the difference between consecutive terms changes sign at each step. It follows that the subsequences of odd, and respectively even, indices are each monotonic, and of opposed monotonicity.

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You can also see it as a fixpoint problem for $f(x) = \sqrt{1+\frac{2}{x}}$ on $[1,2]$ because $$f:[1,2]\rightarrow [1,2] \mbox{ and } |f´(x)| = \left|-\frac{1}{x^2\sqrt{1+\frac{2}{x}}} \right| <= \frac{1}{\sqrt{3}}<1 \mbox{ on } [1,2]$$ So, the limit exists and is the only real solution to $x^3-x-2=0$.

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