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Can someone please help can I write this in the form of a recurrence relation?

$$ \sqrt{5 + \sqrt{6 + \sqrt{7 + ... }}}$$

Thank you

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Let $a_n = \sqrt{ n+4 + \sqrt{n+5 + \sqrt{ n+ 6 + \sqrt{ \ldots } } } } $

Hint: Consider $a_n^2 - a_{n+1}$.

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I've got $$a_{n+1} = a_{n}^{2} - (n+4)$$ but can this be simplified further? :S – user78416 May 28 '13 at 14:34
Isn't that a recurrence relation? So it answered your question. If you want it to be 'simplified', you have to explicitly state what you mean, and what you hope to achieve. – Calvin Lin May 28 '13 at 14:35
Sorry to bother you again, but I was hoping more for a recurrence relation to give the radical sequentially, like: $$x_{1} = \sqrt{5}, x_{2} = \sqrt{5 + \sqrt{6}}, x_{3} = \sqrt{5 + \sqrt{6 + \sqrt{7}}}$$ and so on. With this, if I define $$a_{1} = \sqrt{5}$$ then I get $$a_{2} = (\sqrt{5})^{2} - (2 + 4) = -1$$ Or have I done something stupid? – user78416 May 28 '13 at 14:54
@user78416 The same recurrence relation will not hold, if you have changed the terms. You will need another recurrence relation to use, and there isn't one that immediately pops into my mind. However, I get the sense that you're trying to do something else instead (for example, showing that it is bounded above). If so, you should create another question and state that. – Calvin Lin May 28 '13 at 15:07

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