Evaluation of the 'Odd Harmonic' Continued Fraction How to prove that this continued fraction $1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\cfrac{1/7}{1+\ddots}}}}$ evaluates to
$\displaystyle\frac{2}{\displaystyle 1+\frac{I_1(\frac14)}{I_0(\frac14)}}$?
The recurrence relation of these Besselfunctions is a different one than the recurrence associated with this Continued Fraction.
 A: The continued fraction $\cfrac{1}{1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\cfrac{1/7}{1+\ddots}}}}}$ can be evaluated as follows:
Consider $w_n=\cfrac{1}{1+\cfrac{w_{n+1}}{(2n+1)}}$
Assume $w_n=\cfrac{I_{n+1}}{I_n}$
This results in the recurrence relation ${(1+2n)I_n}={(1+2n)I_{n+1}}+{I_{n+2}}$
Euler's Differential method can be used to find an integral which satisfies this recurrence. Applying Euler's method with $a=1, \alpha=2, b=1, \beta=2, c=1, \gamma=0$ results in a ODE. See Sergey Khruschev's Orthogonal Polynomials and Continued Fractions: From Euler's Point of View for technical details of Euler's Differential Method.
The choice R(x)=x satisfies and results in:
$$w_0=\cfrac{I_1}{I_0}=\cfrac{1}{1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\ddots}}}}=\cfrac{\int_{0}^{1} e^{x/2}\cfrac{x^{1/2}}{(x-1)^{1/2}}dx}{\int_{0}^{1} e^{x/2}\cfrac{1}{x^{1/2}(x-1)^{1/2}}dx}$$
Using the substitution $x=\cfrac{1+cos(\phi)}{2}$ it follows easily that this quotient of integrals equals $\frac{\displaystyle I_0(\frac14)+I_1(\frac14)}{\displaystyle2I_0(\frac14)}$
Taken reciprocals proves that $1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\cfrac{1/7}{1+\ddots}}}} = \frac{\displaystyle 2}{\displaystyle 1+\frac{\displaystyle I_1(\frac14)}{I_0(\frac14)}}$
