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There is a continued fraction for $$\log\left(\frac{1+z}{1-z}\right)$$ given by $$\cfrac{2z}{1-\cfrac{1z^2}{3-\cfrac{4z^2}{\cdots}}}.$$

This was known even to Gauss. See Handbook of Continued Fractions for Special Functions by Annie A.M. Cuyt et al pg 196 or Analytic theory of Continued fractions by H.S. Wall pg 343.

Is there a continued fraction of the same form for the more general $$\log\left(\frac{1+\alpha z}{1-z}\right)?$$

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  • $\begingroup$ The original result is a special case of these, again by Gauss. Your last function doesn't seem to be "close" to a quotient of hypergeometric series, so I wouldn't expect a CF "of the same form". $\endgroup$ – metamorphy Nov 14 '20 at 7:07
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There is $$ \log \left(\frac{a z+1}{1-z}\right)-\frac{(a+1) z}{1-\frac{\frac{1-a}{2} z}{1-\frac{\frac{(a+1)^2}{6 (1-a)} z}{1-\frac{\frac{a^5-a^4-8 a^3-8 a^2-a+1}{3(-a^4-2 a^3+2 a+1)} z}{1-\ldots}}}}. $$ Any formal power series can be expanded into a cfrac of the form $a_0+\operatorname{K}\limits_{n=0}^\infty\frac{a_nz}{1}$, where coefficients $a_n$ can be determined through Hankel matrices composed of coefficients of the formal power series. But one can not always expect a regular pattern in the coefficients $a_n$.

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