Wolfram|Alpha states that the infinite continued fraction $$\cfrac{1}{0+\cfrac{1}{0+\cdots}}=0.$$ Assuming $[0;0,0,\ldots]$ exists implies that the continued fraction is $1$, since $x=\dfrac{1}{0+\cfrac{1}{0+\cdots}}=\dfrac{1}{x}$ implies $x=1$ (ignoring the negative value). This, along with the divide by zero error, suggests W|A is wrong.

Is this an error on W|A's part? If not, is this just a convention, and is there a similar convention for $$\cfrac{0}{0+\cfrac{0}{0+\cdots}}?$$

  • $\begingroup$ Did you mean to write $1$ or $0$ in the numerators? $\endgroup$ – String Oct 5 '13 at 23:51
  • $\begingroup$ I accidentally wrote 1 as 0 in my second equation (the equation with x); this is fixed now. $\endgroup$ – Daniel Hendrycks Oct 6 '13 at 0:07
  • 3
    $\begingroup$ You are correct that if the expression means anything then it equals its own reciprocal so it has to be $1$ or $-1$. But I think if you go to the definition of simple continued fraction you find it doesn't mean anything. $\endgroup$ – Gerry Myerson Oct 6 '13 at 0:39

As Gerry indicates in his comment: the definition of an infinite continued fraction is the limit of its finite truncations. Here none of the truncations are well-defined, so certainly neither is the limit.


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