The wikipedia page on Generalized Continued Fractions gives this equation. $$ \log \left( 1+\frac{x}{y} \right) = \cfrac{x} {y+\cfrac{1x} {2+\cfrac{1x} {3y+\cfrac{2x} {2+\cfrac{2x} {5y+\cfrac{3x} {2+\ddots}}}}}} = \cfrac{2x} {2y+x-\cfrac{(1x)^2} {3(2y+x)-\cfrac{(2x)^2} {5(2y+x)-\cfrac{(3x)^2} {7(2y+x)-\ddots}}}} $$
Is there an inductive way to prove the equality between the two continued fractions?
This is what [2] calls an even canonical contraction: The sequence of approximants of the contracted continued fraction corresponds to the sequence of even-indexed approximants of the original continued fraction.
In this example, we have: $$\begin{align} b_0 + \dfrac{a_1}{b_1+\dfrac{a_2}{\ddots+\dfrac{a_{2n}}{b_{2n}}}} = d_0 + \dfrac{c_1}{d_1+\dfrac{c_2}{\ddots+\dfrac{c_n}{d_n}}} \end{align}$$ where for $k\geq1$ $$\begin{align} b_0 &= 0 & a_1 &= x \\ b_{2k-1} &= (2k-1)y & a_{2k} &= kx \\ b_{2k} &= 2 & a_{2k+1} &= kx \end{align}$$ and after even canonical contraction ($m\geq2$): $$\begin{align} d_0 &= b_0 = 0 & c_1 &= a_1 b_2 = 2x \\ d_1 &= a_2 + b_1 b_2 & c_m &= -\frac{a_{2m-2}a_{2m-1}b_{2m}}{b_{2m-2}} \\ &= x + 2y & &= -(m-1)^2 x^2 \\ d_m &= a_{2m} + b_{2m-1}b_{2m} + \frac{a_{2m-1}b_{2m}}{b_{2m-2}} \\ &= (2m-1)(2y + x) \end{align}$$ The formulae are from section 1.5 of [2]. A proof can be found in section 2.4 of [1]. It is explicit rather than inductive and builds on more general theorems for almost arbitrary contractions and extensions of approximant sequences of continued fractions. You will have to apply equivalence transformations to match the formula given in [2].
[1] Jones and Thron: Continued Fractions - Analytic Theory and Applications. In: Rota (ed.): Encyclopedia of Mathematics and its Applications, Vol. 11, Section: Analysis. Addison-Wesley 1980.
[2] Cuyt, Petersen, Verdonk, Waadeland, Jones: Handbook of Continued Fractions for Special Functions. Springer 2008, e-ISBN: 978-1-4020-6949-9
