# Unsure about infinite continued fraction

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? $:)$

• Take a look at this mathworld.wolfram.com/ContinuedFractionConstant.html . I believe some continued fractions aren't known to be expressed in simple forms. – Jam Aug 11 '14 at 13:54
• The standard continued fraction of such a constant does not look to be (eventually) periodic like in the case of quadratic irrationals (Lagrange theorem, faculty.plattsburgh.edu/sam.northshield/09-0507.pdf) are you really sure the integers $p,q,k,n$ exist? – Jack D'Aurizio Aug 11 '14 at 14:00
• For istance, I get $[0; 2, 1, 1, 1, 2, 1, 2, 10, 2, 2, 66,1,1,13,\ldots]$ as the standard continued fraction for $a$. – Jack D'Aurizio Aug 11 '14 at 14:02
• It is definitely not quadratic since the continued fraction for a quadratic number is repeating – Alice Ryhl Aug 11 '14 at 15:10
• @Darksonn Only the standard continued fraction need repeat. The standard continued fraction has its partial numerators all 1s. This does not rule out a nonrepeating, nonstandard continued fraction such as the one OP is asking about. – MJD Aug 11 '14 at 16:51

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.