Unsure about infinite continued fraction

Question

How do you/is it possible to express $a=\cfrac{1}{2+\cfrac{3}{4+\cfrac{5}{6+\cdots}}}$ in the form $\frac{p}{q}(k+\sqrt{n})$? I'm still in high school, so I'm not familiar with especially sophisticated approaches for evaluating infinitely continued fractions - I've been able to set up a recurrence relation for everything I've encountered so far, but after writing it out as $a(x)=\cfrac{x}{x+1+\cfrac{x+2}{x+3+\cdots}}$, I'm fairly sure that approach isn't going to work. I'm also not sure how to represent this in continued fraction notation, as each nested fraction has a unique numerator. It clearly converges to something, so can anyone point me in the right direction? $:)$

Answer 1

In view of the fact that the standard continued fraction expansion of $e$ is $$ e=2+\frac1{1+}\frac1{2+}\frac1{1+}\frac1{1+}\frac1{4+}\frac1{1+}\frac1{1+}\frac1{6+}\cdots\,, $$ I think you are mistaken in thinking that your number might even be algebraic, much less quadratic.

Answer 2

The continued fraction $0+\cfrac{1}{2+\cfrac{3}{4+\cfrac{5}{6+\cfrac{7}{8+\cfrac{9}{10+\ddots}}}}}$ can be evaluated using Euler's Differential method. Applying this method with $a=1, \alpha=2, b=2, \beta=2, c=1, \gamma=0$ results in a ODE. See Sergey Khruschev's Orthogonal Polynomials and Continued Fractions: From Euler's Point of View for technical details.
Solving the ODE results in $0+\cfrac{1}{2+\cfrac{3}{4+\cfrac{5}{6+\cfrac{7}{8+\cfrac{9}{10+\ddots}}}}}=\cfrac{\int_{0}^{1} e^{x/2}x^{1/2}dx}{\int_{0}^{1} e^{x/2}{x^{-1/2}}dx}=0.379732...$

Helas, this result can not be expressed easily in common functions. Of course, it can be expressed as the quotient of two series.

The interesting 'mirrored' Continued Fraction is $1+\cfrac{2}{3+\cfrac{4}{5+\cfrac{6}{7+\cfrac{8}{9+\cfrac{10}{11+\ddots}}}}}$

And this CF can also be evaluated using Euler's Differential method.

The interesting result is that $ 1+\cfrac{2}{3+\cfrac{4}{5+\cfrac{6}{7+\cfrac{8}{9+\cfrac{10}{11+\ddots}}}}}=\cfrac{1}{\sqrt{e} -1}=1.54149...$