Some of Ramanujan’s continued fractions for pi

The digits of pi go on forever apparently with no discernible pattern. However, there are beautifully simple patterns in its (ironically) non-simple continued fraction expansions. Examples are,

\pi = \cfrac{4}{1 + \cfrac{1^2}{2 + \cfrac{3^2}{2 + \cfrac{5^2}{2 + \ddots} }}}

known by Lord Brouncker (1620-1684), and,

\pi = 3 + \cfrac{1^2}{6 + \cfrac{3^2}{6 + \cfrac{5^2}{6 + \ddots} }}

One can see the affinity between the two.  They in fact belong to the same family.  Given complex numbers {n, x} with Re[x] > 0, then,

F(n,x) = \Large{\frac{\Gamma\left(\frac{x+n+1}{4} \right)\Gamma\left(\frac{x-n+1}{4} \right) }{\Gamma\left(\frac{x+n+3}{4} \right)\Gamma\left(\frac{x-n+3}{4} \right)}} = \cfrac{4}{x + \cfrac{1^2-n^2}{2x + \cfrac{3^2-n^2}{2x + \cfrac{5^2-n^2}{2x + \ddots} }}}

where \Gamma(\tau) is the gamma function.  This is Entry 25 in Ramanujan’s Second Notebook (Chapter 12) though this result was also known by Euler. For the case = 0, the continued fraction assumes the form of the examples and the function simplifies as,

F(0,x) = \frac{\Gamma^2\left(\frac{x+1}{4}\right)}{\Gamma^2\left(\frac{x+3}{4}\right)}

For x an odd integer, then (0, x) is a rational multiple of \pi or 1/\pi.  Specifically, for x = {1, 3, 5, 7, 9}, we have,

F(0,x) = \pi,\; \frac{4}{\pi}, \frac{\pi}{4}, \frac{16}{9\pi}, \frac{9\pi}{64}

More generally,

F(0,4m+1) = \big(\frac{(2m)!}{m!^2}\big)^2\, \frac{\pi}{2^{4m}}

F(0,4m+3) = \big(\frac{m!^2}{(2m+1)!}\big)^2\, \frac{2^{4m+2}}{\pi}

See p. 178 of Annie Cuyt’s HandBook of Continued Fractions for Special Functions.  For more examples, see also this article Ramanujan’s Continued Fractions, Apery’s Constant, and more.

One response to this post.

  1. That’s a beautiful continued fraction


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