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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?

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    $\begingroup$ proof of the continued fraction of $e$. $\endgroup$
    – Dando18
    Jul 7, 2017 at 13:31
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    $\begingroup$ There is a way to convert power series into continued fractions. $\endgroup$
    – Simply Beautiful Art
    Jul 7, 2017 at 13:32
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    $\begingroup$ also, I feel this question might answer yours $\endgroup$
    – Dando18
    Jul 7, 2017 at 13:40
  • $\begingroup$ You might be interested in Thiele expansion (see e.g. this), which is the CF analog of Taylor expansion. Another useful source of continued fractions are linear three-term recurrences: they can be suitably rearranged into continued fractions that converge to ratios of functions that are the solutions of the three-term recurrence, via Pincherle's theorem. $\endgroup$
    – J. M. ain't a mathematician
    Nov 14, 2017 at 15:09

2 Answers 2

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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 $\pi$.

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    $\begingroup$ The "after some algebra, you'll be able to derive..." part is the crux of the matter. If one directly uses the Euler continued fraction formula + a simple equivalence transformation to change the denominators, one ends up with: $$\arctan x=\frac{x}{1+\frac{x^2}{3-x^2+\frac{(3x)^2}{5-3x^2+\frac{(5x)^2}{7-5x^2+\cdots}}}}$$But I am not aware of any simple transformation techniques to turn this into the form you provide $\endgroup$
    – FShrike
    Aug 20, 2022 at 10:43
  • $\begingroup$ Sorry, but $\arctan(x) = x - x^3/3 + x^5/5 - \cdots$ with $(-1)^n$ in the numerator instead of $(-2)^n$. $\endgroup$
    – Somos
    Aug 20, 2022 at 20:29
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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.

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