# Euler's continued fraction formula

In the analytic theory of continued fractions, Euler's continued fraction formula is an identity connecting a certain very general infinite series with an infinite continued fraction. First published in 1748, it was at first regarded as a simple identity connecting a finite sum with a finite continued fraction in such a way that the extension to the infinite case was immediately apparent.[1] Today it is more fully appreciated as a useful tool in analytic attacks on the general convergence problem for infinite continued fractions with complex elements.

## The original formula

Euler derived the formula as connecting a finite sum of products with a finite continued fraction.

${\displaystyle a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+\cdots +a_{0}a_{1}a_{2}\cdots a_{n}={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{1+a_{1}-{\cfrac {a_{2}}{1+a_{2}-{\cfrac {\ddots }{\ddots {\cfrac {a_{n-1}}{1+a_{n-1}-{\cfrac {a_{n}}{1+a_{n}}}}}}}}}}}}}\,}$

The identity is easily established by induction on n, and is therefore applicable in the limit: if the expression on the left is extended to represent a convergent infinite series, the expression on the right can also be extended to represent a convergent infinite continued fraction.

## Euler's formula

If ri are complex numbers and x is defined by

${\displaystyle x=1+\sum _{i=1}^{\infty }r_{1}r_{2}\cdots r_{i}=1+\sum _{i=1}^{\infty }\left(\prod _{j=1}^{i}r_{j}\right)\,,}$

then this equality can be proved by induction

${\displaystyle x={\cfrac {1}{1-{\cfrac {r_{1}}{1+r_{1}-{\cfrac {r_{2}}{1+r_{2}-{\cfrac {r_{3}}{1+r_{3}-\ddots }}}}}}}}\,}$.

Here equality is to be understood as equivalence, in the sense that the n'th convergent of each continued fraction is equal to the n'th partial sum of the series shown above. So if the series shown is convergent – or uniformly convergent, when the ri's are functions of some complex variable z – then the continued fractions also converge, or converge uniformly.[2]

## Examples

### The exponential function

The exponential function ez is an entire function with a power series expansion that converges uniformly on every bounded domain in the complex plane.

${\displaystyle e^{z}=1+\sum _{n=1}^{\infty }{\frac {z^{n}}{n!}}=1+\sum _{n=1}^{\infty }\left(\prod _{j=1}^{n}{\frac {z}{j}}\right)\,}$

The application of Euler's continued fraction formula is straightforward:

${\displaystyle e^{z}={\cfrac {1}{1-{\cfrac {z}{1+z-{\cfrac {{\frac {1}{2}}z}{1+{\frac {1}{2}}z-{\cfrac {{\frac {1}{3}}z}{1+{\frac {1}{3}}z-{\cfrac {{\frac {1}{4}}z}{1+{\frac {1}{4}}z-\ddots }}}}}}}}}}.\,}$

Applying an equivalence transformation that consists of clearing the fractions this example is simplified to

${\displaystyle e^{z}={\cfrac {1}{1-{\cfrac {z}{1+z-{\cfrac {z}{2+z-{\cfrac {2z}{3+z-{\cfrac {3z}{4+z-\ddots }}}}}}}}}}\,}$

and we can be certain that this continued fraction converges uniformly on every bounded domain in the complex plane because it is equivalent to the power series for ez.

### The natural logarithm

The Taylor series for the principal branch of the natural logarithm in the neighborhood of z = 1 is well known. Recognizing that log(a/b) = log(a) - log(b), the following series is easily derived:

${\displaystyle \log {\frac {1+z}{1-z}}=2\left(z+{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}+\cdots \right)=2\sum _{n=0}^{\infty }{\frac {z^{2n+1}}{2n+1}}.\,}$

This series converges when |z| < 1 and can also be expressed as a sum of products:[3]

{\displaystyle {\begin{aligned}\log {\frac {1+z}{1-z}}&=2z\left[1+{\frac {z^{2}}{3}}+{\frac {z^{4}}{5}}+\cdots \right]\\[8pt]&=2z\left[1+{\frac {z^{2}}{3}}+\left({\frac {z^{2}}{3}}\right){\frac {z^{2}}{5/3}}+\left({\frac {z^{2}}{3}}\right)\left({\frac {z^{2}}{5/3}}\right){\frac {z^{2}}{7/5}}+\cdots \right]\end{aligned}}}

Applying Euler's continued fraction formula to this expression shows that

${\displaystyle \log {\frac {1+z}{1-z}}={\cfrac {2z}{1-{\cfrac {{\frac {1}{3}}z^{2}}{1+{\frac {1}{3}}z^{2}-{\cfrac {{\frac {3}{5}}z^{2}}{1+{\frac {3}{5}}z^{2}-{\cfrac {{\frac {5}{7}}z^{2}}{1+{\frac {5}{7}}z^{2}-{\cfrac {{\frac {7}{9}}z^{2}}{1+{\frac {7}{9}}z^{2}-\ddots }}}}}}}}}}\,}$

and using an equivalence transformation to clear all the fractions results in

${\displaystyle \log {\frac {1+z}{1-z}}={\cfrac {2z}{1-{\cfrac {z^{2}}{z^{2}+3-{\cfrac {(3z)^{2}}{3z^{2}+5-{\cfrac {(5z)^{2}}{5z^{2}+7-{\cfrac {(7z)^{2}}{7z^{2}+9-\ddots }}}}}}}}}}.\,}$

This continued fraction converges when |z| < 1 because it is equivalent to the series from which it was derived.[3]

### A continued fraction for π

We can use the previous example involving the principal branch of the natural logarithm function to construct a continued fraction representation of π. First we note that

${\displaystyle {\frac {1+i}{1-i}}=i\quad \Rightarrow \quad \log {\frac {1+i}{1-i}}={\frac {i\pi }{2}}.\,}$

Setting z = i in the previous result, and remembering that i2 = −1, we obtain immediately

${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}}.\,}$