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Let $F(n)$ denote a infinite continued fraction of form such that:

$$F(n) = \cfrac{1}{2n + \cfrac{1}{2n + \cfrac{1}{2n + \cfrac{1}{2n + \cfrac{1}{\dots}}}}}$$

Consider the following equation:

$$\sqrt{n^2+1}-n = F(n), n \in \mathbb{N}_{>0}$$

Could this be easily proven? It seems to be correct, but I have no idea where to start...

Any help will be much appreciated!

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  • $\begingroup$ Can you see anything in the continued fraction expression that can be substituted by $F(n)$? $\endgroup$
    – Daniel R
    Aug 8, 2013 at 19:55

1 Answer 1

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Hint:

Try $$F(n) = \frac{1}{2n + F(n)}$$

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  • $\begingroup$ Sweet... Thanks... +1 $\endgroup$
    – JohnWO
    Aug 8, 2013 at 19:56
  • $\begingroup$ No problem, glad I could help! :) $\endgroup$ Aug 8, 2013 at 19:56

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