# Is the infinite continued fraction $[0;0,0,\ldots]=0$?

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 $$\cfrac{0}{0+\cfrac{0}{0+\cdots}}?$$

• Did you mean to write $1$ or $0$ in the numerators? – String Oct 5 '13 at 23:51
• I accidentally wrote 1 as 0 in my second equation (the equation with x); this is fixed now. – Daniel Hendrycks Oct 6 '13 at 0:07
• 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. – Gerry Myerson Oct 6 '13 at 0:39