My exploration to this problem $x_{1}=1,x_{n+1}=\frac{x_{n}}{2}+\frac{1}{x_{n}},n\geq2$ :
I found it in book: The Elements of Cantor Sets With Applications by Robert W. Vallin at page 75 ,this book says it converges.
Then, I collect numerical data that suggest this sequence converges to $\sqrt{2}$, since for $n \geq 2$ it is decreasing and I think its infimum equals $\sqrt{2}$.
But how to prove this rigorously?
I tried mathematical induction but doesn't work.
Applying the inequality $$a^2+b^2\ge 2ab$$ gives $${a_n\over 2}+{1\over a_n}\ge \sqrt{2}$$ Thus $a_n\ge \sqrt{2}$ for $n\ge 2.$ Next for $n\ge 2$ we get $$a_{n+1}-a_n={1\over a_n}-{a_n\over 2}= -{a_n^2-2\over 2a_n}\le 0$$ A much better solution : for $n\ge 2$ $$a_{n+1}-\sqrt{2}={a_n^2-2\sqrt{2}a_n+2\over {2}a_n} ={(a_n-\sqrt{2})^2\over 2a_n}\le {a_n-\sqrt{2}\over 2}$$ Hence $$0\le a_n-\sqrt{2}\le {1\over 2^{n-1}}a_2={3\over 2^n}$$
This is a special case of a more general problem: Convergence of the sequence $\displaystyle a_{n+1} = \frac{1}{p}( (p-1)a_n+\frac{a}{a_n^{p-1}})$
See above if interested.
Good Luck !
