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I can't find anywhere via googling; is there some sort of $\sum$ like notation for infinite continued fractions? In other words, for a sum we do this:

$$ 1+x+x^2+x^3+... = \sum_{n=0}^\infty x^n $$

Easy, this notation is well known. Now, for a continued fraction:

$$ \pi= 3 + \dfrac{1}{6+\dfrac{9}{6+\dfrac{25}{6+\dfrac{49}{6+...}}}} $$

This could be defined recursively, i.e. $\pi = F_0$ where:

$$ F_n= \begin{cases} 3 + \dfrac{(2n+1)^n}{F_{n+1}} & n = 0\\ 6 + \dfrac{(2n+1)^2}{F_{n+1}} & n \geq 1 \\ \end{cases} $$

But this notation lacks the elegance of the $\sum$ example above. Even the notation for simple continued fractions isn't all that much of an improvement, i.e.

$$ \sqrt{2} = 1 + \dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+...}}} = [1; 2, 2, 2, 2, ...] $$

Is there a better way?

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2 Answers 2

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Wikipedia cites a notation by Gauss: $$ x = a_0 + \dfrac{1}{a_1 + \dfrac{1}{a_2 + \dfrac{1}{a_3}}} $$ would be written: $$ x = a_0 + \mathop{\mathrm{K}}_{k = 1}^3 \frac{1}{a_k} $$

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  • $\begingroup$ So then $\pi = 3 + \mathop{\mathrm{K}}_{k = 0}^\infty \frac{(2k+1)^2}{6}$? $\endgroup$
    – durron597
    Feb 25, 2014 at 14:49
  • $\begingroup$ @durron597, that seems the logical extension of the notation. $\endgroup$
    – vonbrand
    Feb 25, 2014 at 14:51
  • $\begingroup$ However, it is pretty confusing since $\frac{(2k+1)^2}{6}\neq\frac{(4k+2)^2}{24}$. $\endgroup$
    – yo'
    Feb 26, 2014 at 9:42
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I believe that the notation in Hatcher Topology of Numbers is the best notation. No sooner had I seen it than I got love with the notation. The notation is $$\pi=3 +{}^{1}\mkern-10mu\nearrow\mkern-8mu{}_{\normalsize 7} +{}^{1}\mkern-10mu\nearrow\mkern-8mu{}_{\normalsize 15} +{}^{1}\mkern-10mu\nearrow\mkern-8mu{}_{\normalsize 1} +{}^{1}\mkern-10mu\nearrow\mkern-8mu{}_{\normalsize 292} +{}^{1}\mkern-10mu\nearrow\mkern-8mu{}_{\normalsize 1} +{}^{1}\mkern-10mu\nearrow\mkern-8mu{}_{\normalsize 1} +\ldots$$

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