Convergence problem

An infinite periodic continued fraction is a continued fraction of the form

${\displaystyle x={\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+{\cfrac {\ddots }{\quad \ddots \quad b_{k-1}+{\cfrac {a_{k}}{b_{k}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+\ddots }}}}}}}}}}}}\,}$

where k ≥ 1, the sequence of partial numerators {a₁, a₂, a₃, ..., a_k} contains no values equal to zero, and the partial numerators {a₁, a₂, a₃, ..., a_k} and partial denominators {b₁, b₂, b₃, ..., b_k} repeat over and over again, ad infinitum.

By applying the theory of linear fractional transformations to

${\displaystyle s(w)={\frac {A_{k-1}w+A_{k}}{B_{k-1}w+B_{k}}}\,}$

where A_k-1, B_k-1, A_k, and B_k are the numerators and denominators of the k-1st and kth convergents of the infinite periodic continued fraction x, it can be shown that x converges to one of the fixed points of s(w) if it converges at all. Specifically, let r₁ and r₂ be the roots of the quadratic equation

${\displaystyle B_{k-1}w^{2}+(B_{k}-A_{k-1})w-A_{k}=0.\,}$

These roots are the fixed points of s(w). If r₁ and r₂ are finite then the infinite periodic continued fraction x converges if and only if

1. the two roots are equal; or
2. the k-1st convergent is closer to r₁ than it is to r₂, and none of the first k convergents equal r₂.

If the denominator B_k-1 is equal to zero then an infinite number of the denominators B_nk-1 also vanish, and the continued fraction does not converge to a finite value. And when the two roots r₁ and r₂ are equidistant from the k-1st convergent – or when r₁ is closer to the k-1st convergent than r₂ is, but one of the first k convergents equals r₂ – the continued fraction x diverges by oscillation.[1][2][3]

If the period of a continued fraction is 1; that is, if

${\displaystyle x={\underset {1}{\overset {\infty }{\mathrm {K} }}}{\frac {a}{b}},\,}$

where b ≠ 0, we can obtain a very strong result. First, by applying an equivalence transformation we see that x converges if and only if

${\displaystyle y=1+{\underset {1}{\overset {\infty }{\mathrm {K} }}}{\frac {z}{1}}\qquad \left(z={\frac {a}{b^{2}}}\right)\,}$

converges. Then, by applying the more general result obtained above it can be shown that

${\displaystyle y=1+{\cfrac {z}{1+{\cfrac {z}{1+{\cfrac {z}{1+\ddots }}}}}}\,}$

converges for every complex number z except when z is a negative real number and z < −¼. Moreover, this continued fraction y converges to the particular value of

${\displaystyle y={\frac {1}{2}}\left(1\pm {\sqrt {4z+1}}\right)\,}$

that has the larger absolute value (except when z is real and z < −¼, in which case the two fixed points of the LFT generating y have equal moduli and y diverges by oscillation).

By applying another equivalence transformation the condition that guarantees convergence of

${\displaystyle x={\underset {1}{\overset {\infty }{\mathrm {K} }}}{\frac {1}{z}}={\cfrac {1}{z+{\cfrac {1}{z+{\cfrac {1}{z+\ddots }}}}}}\,}$

can also be determined. Since a simple equivalence transformation shows that

${\displaystyle x={\cfrac {z^{-1}}{1+{\cfrac {z^{-2}}{1+{\cfrac {z^{-2}}{1+\ddots }}}}}}\,}$

whenever z ≠ 0, the preceding result for the continued fraction y can be restated for x. The infinite periodic continued fraction

${\displaystyle x={\underset {1}{\overset {\infty }{\mathrm {K} }}}{\frac {1}{z}}}$

converges if and only if z² is not a real number lying in the interval −4 < z² ≤ 0 – or, equivalently, x converges if and only if z ≠ 0 and z is not a pure imaginary number with imaginary part between -2 and 2. (Not including either endpoint)

By applying the fundamental inequalities to the continued fraction

${\displaystyle x={\cfrac {1}{1+{\cfrac {a_{2}}{1+{\cfrac {a_{3}}{1+{\cfrac {a_{4}}{1+\ddots }}}}}}}}\,}$

it can be shown that the following statements hold if |a_i| ≤ ¼ for the partial numerators a_i, i = 2, 3, 4, ...

• The continued fraction x converges to a finite value, and converges uniformly if the partial numerators a_i are complex variables.[4]
• The value of x and of each of its convergents x_i lies in the circular domain of radius 2/3 centered on the point z = 4/3; that is, in the region defined by

${\displaystyle \Omega =\lbrace z:|z-4/3|\leq 2/3\rbrace .\,}$[5]

• The radius ¼ is the largest radius over which x can be shown to converge without exception, and the region Ω is the smallest image space that contains all possible values of the continued fraction x.[5]

The proof of the first statement, by Julius Worpitzky in 1865, is apparently the oldest published proof that a continued fraction with complex elements actually converges.[6]

Because the proof of Worpitzky's theorem employs Euler's continued fraction formula to construct an infinite series that is equivalent to the continued fraction x, and the series so constructed is absolutely convergent, the Weierstrass M-test can be applied to a modified version of x. If

${\displaystyle f(z)={\cfrac {1}{1+{\cfrac {c_{2}z}{1+{\cfrac {c_{3}z}{1+{\cfrac {c_{4}z}{1+\ddots }}}}}}}}\,}$

and a positive real number M exists such that |c_i| ≤ M (i = 2, 3, 4, ...), then the sequence of convergents {f_i(z)} converges uniformly when

${\displaystyle |z|<{\frac {1}{4M}}\,}$

and f(z) is analytic on that open disk.