I'd like to know the general term of the following recurrence $$a_{n+1}=a_n + \frac{\alpha}{a_n}, $$ where $\alpha>0$ and $a_0=M>0$. That is, ideally I'd like to find the function $g$ such that $a_n = g(n;\alpha,M)$. Any idea? Many thanks !
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Here is an idea that may not work for your case but nails it if $$a_{n+1} = \frac12\left( a_n+ \frac{\alpha}{a_n}\right)$$ Let $a_n = \tanh z$, then you will find something like $a_{n+1} = \tanh (2z)$ and the closed form becomes easy. (If $\alpha < 0$ use $\tan z$ instead.) The idea does not work unchanged, but may provide a starting point. – Mark Fischler Mar 24 '17 at 01:14
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With $,a_n = \sqrt{\alpha} ,b_n,$ the recurrence becomes $b_{n+1} = b_n + \cfrac{1}{b_n},$. Maybe relevant: Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$. – dxiv Mar 24 '17 at 01:31
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1Many thanks! Yes @dxiv that discussion was very helpful – Luke Mar 25 '17 at 01:27