How to solve this nonlinear difference equation $a_{n+1} = 2a_n + \frac{1}{a_n}$, $a_1 = 1$?

Question

How to solve this nonlinear difference equation $$a_{n+1} = 2a_n + \frac{1}{a_n},\quad a_1 = 1.$$

Answer 1

Clearly, all the $a_n$'s are positive. So, $a_{n+1}>2a_n$ for every $n$, this implies that $a_n\ge 2^{n-1}$ for every $n$.

Plugging this in the difference equation, shows that $$0<\frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=\frac{1}{2^{n+1}a_n}\le \frac{1}{2^{2n}}$$ Adding we conclude that there exists a number $k\in(1/2,5/6) $ such that $\lim\limits_{n\to\infty}a_n2^{-n}=k$.