How to solve this nonlinear difference equation $$a_{n+1} = 2a_n + \frac{1}{a_n},\quad a_1 = 1.$$
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Are you looking for a closed form expression for $a_n$, or is solving this recurrence a smaller step in a larger problem? – JimmyK4542 Aug 05 '15 at 00:41
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I just want to know if there is an expression, maybe not closed-form, for $a_n$. – wayne Aug 05 '15 at 01:38
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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 $\lim\limits_{n\to\infty}a_n2^{-n}=k$.
Omran Kouba
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