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

Question

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!

Answer 1

Hint:

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