<s>[INST] Is the sequence defined by the recurrence $ a _ { n + 2 } = \frac 1 { a _ { n + 1 } } + \frac 1 { a _ n } $ convergent? 
Let $ a _ 0 = 1 $, $ a _ 1 = 1 $ and $ a _ { n + 2 } = \frac 1 { a _ { n + 1 } } + \frac 1 { a _ n } $ for every natural number $ n $. How can I prove that this sequence is convergent?


I know that if it's convergent, it converges to $ \sqrt 2 $ since if $ \lim \limits _ { n \to \infty } a _ n = a $ then:
$$ \lim _ { n \to \infty } \left ( a _ { n + 2 } - \frac 1 { a _ { n + 1 } } - \frac 1 { a _ n } \right) = a - \frac 2 a = 0 \text ; $$
$$ \therefore \quad a ^ 2 = 2 \text . $$
Now it's easy to see that every $ a _ n $ is positive, so $ a \ge 0 $ and thus $ a = \sqrt 2 $.

Assuming the sequence is convergent, I can calculate an estimation of the rate of convergence too. Let $ \epsilon _ n := a _ n - \sqrt 2 $. We have:
$$ \epsilon _ { n + 2 } = \frac 1 { a _ { n + 1 } } - \frac 1 { \sqrt 2 } + \frac 1 { a _ n } - \frac 1 { \sqrt 2 } = - \frac { a _ { n + 1 } - \sqrt 2 } { \sqrt 2 a _ { n + 1 } } - \frac { a _ n - \sqrt 2 } { \sqrt 2 a _ n } = - \frac { \epsilon _ { n + 1 } } { \sqrt 2 a _ { n + 1 } } - \frac { \epsilon _ n } { \sqrt 2 a _ n } \text . $$
Now because $ a _ n \sim \sqrt 2 + \epsilon _ n $ and $ \lim \limits _ { n \to \infty } \epsilon _ n = 0 $, therefore from the above equation:
$$ \epsilon _ { n + 2 } \lesssim - \frac { \epsilon _ { n + 1 } + \epsilon _ n } 2 \text , $$
which yields $ \epsilon _ n \lesssim \alpha \left ( \frac { - 1 - \sqrt 7 i } 4 \right) ^ n + \beta \left( \frac { - 1 + \sqrt 7 i } 4 \right) ^ n $ for some complex constants $ \alpha $ and $ \beta $, using induction on $ n $. Equivalently, we have $ \epsilon _ n \lesssim \left( \frac 1 { \sqrt 2 } \right) ^ n \bigl( A \cos ( n \theta ) + B \sin ( n \theta ) \bigr) $ for $ \theta = \arctan \frac { \sqrt 7 } 4 $ and some real constants $ A $ and $ B $, since $ \left| \frac { - 1 \pm \sqrt 7 i } 4 \right| = \frac 1 { \sqrt 2 } $ and $ \arg \frac { - 1 \pm \sqrt 7 i } 4 = \pi \mp \theta $. Hence we get the rough estimation $ | \epsilon _ n | \lesssim C 2 ^ { - \frac n 2 } $ for some real constant $ C $, and $ \frac 1 { \sqrt 2 } $ is a good guess for the rate of convergence.
(Edit: Thanks to Alex Ravsky for the confirming graphs in his answer.)

Edit (some more of my thoughts):
Let $ b _ n := \min \left\{ a _ n , a _ { n + 1 } , \frac 2 { a _ n } , \frac 2 { a _ { n + 1 } } \right\} $. It's easy to see that $ b _ n \le a _ n \le \frac 2 { b _ n } $ and $ b _ n \le a _ { n + 1 } \le \frac 2 { b _ n } $. Now using induction we can prove that $ b _ n \le a _ { n + m } \le \frac 2 { b _ n } $. Especially, $ a _ { n + 2 } \ge b _ n $ and $ \frac 2 { a _ { n + 2 } } \ge b _ n $ which yields $ b _ { n + 1 } \ge b _ n $. The problem can be solved if I show that the sequence $ ( b _ n ) _ { n = 0 } ^ \infty $ increases to $ \sqrt 2 $.
 [/INST] Graphs, illustrating asymptotic behavior of the sequence $\{a_n\}$. The graphs suggest that $$(a_n-\sqrt{2})\sqrt{2}^n=O(1).$$ 


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