How does one come up with a Continued Fraction? All over the place on Wikipedia, I see a bunch of identities related to continued fractions, like
$$\arctan x=\cfrac{x}{1+\cfrac{x^2}{3+\cfrac{4x^2}{5+\cfrac{9x^2}{7+...}}}}$$
or
$$\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+...}}}}$$
How does one verify these? I haven't even managed to prove a single one of them yet.
Furthermore, aside from proving them, how does one derive them? That is, how does one come up with something like this?
I know how to evaluate simpler continued fractions, like
$$\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+...}}}$$
But I don't know how to evaluate anything where the terms on the inside follow some other sort of sequence.
Can anyone provide a proof (or, preferably, a derivation) of one of the two identities above, or some other continued fraction identity? Can anybody give me some way to go about these types of problems?
 A: Note we have Euler's continued fraction formula, which states:
$$a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+\cdots +a_{0}a_{1}a_{2}\cdots a_{n}\\=\cfrac  {a_{0}}{1-{\cfrac  {a_{1}}{1+a_{1}-{\cfrac  {a_{2}}{1+a_{2}-{\cfrac  {\ddots }{\ddots {\cfrac  {a_{{n-1}}}{1+a_{{n-1}}-{\cfrac  {a_{n}}{1+a_{n}}}}}}}}}}}}$$
This may easily be proven by induction and lends itself for easy conversion of a series into a fraction.  For example, we know that:
$$\arctan(x)=\sum_{n=0}^\infty\frac{(-2)^nx^{2n+1}}{2n+1}$$
Set $a_0=x$ and inductively,
$$a_{n+1}=\frac{(-1)^{n+1}x^{2n+3}}{2n+3}\prod_{k=0}^n(a_k)^{-1}=-\frac{2n+1}{2n+3}x^2$$
And then after some algebra, you'll be able to derive
$$\arctan(x)=\cfrac{x}{1+\cfrac{x^2}{3+\cfrac{4x^2}{5+\cfrac{9x^2}{7+\ddots}}}}$$
Plug $x=1$ in and do some more algebra and you'll end up with the formula for $\pi$.
A: Try 
http://dainoequinoziale.it/resources/sassolini/pifrazionecontinua.pdf
It is in Italian, but it explains in the simplest possible way how to go from Lord Brouncker continued fraction to the Gregory-Leibniz series to calculate Pi. 
