I was attempting to evaluate the following infinite fraction:


So I let $x=\frac{-1}{\frac{-1}{\frac{-1}{\dots}}}$, thus $x=\frac{-1}{x}$ and we arrive at $x^2=-1$, so $x=\pm i$.

Is this correct?

  • 3
    $\begingroup$ You need to define precisely the meaning of the dots. It is usually defined as the limit of the sequence $x_{n+1}=\frac{-1}{x_n}$, provided this limit exists. What you proved is: IF the limit exists, then it is either $i$ or $-1$. $\endgroup$ – Taladris Oct 24 '17 at 7:42
  • $\begingroup$ So your infinite fraction is an interesting way of writing down the limit of the sequence $-1,1,-1,1,-1,\dots$ which doesn't exist $\endgroup$ – user340297 Oct 24 '17 at 7:48
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    $\begingroup$ No, because the limit of a real sequence, if existent, is unique and cannot be an element of $\mathbb{C}\setminus\mathbb{R}$. The map $x\to-\frac{1}{x}$ has no real fixed points. $\endgroup$ – Jack D'Aurizio Oct 24 '17 at 10:54
  • $\begingroup$ @Taladris You mean $-i$. $\endgroup$ – Simply Beautiful Art Oct 24 '17 at 21:29
  • $\begingroup$ To add onto @JackD'Aurizio even with an initial complex seed, it may not converge. $\endgroup$ – Simply Beautiful Art Oct 24 '17 at 21:30

All of these and similar "tricks" for evaluating infinite expressions are really arguments of the form,

"If the expression converges to some number $x$, then the following must be true about $x$..."

So you have indeed proven that if your expression converges, it does so to either $i$ or $-i$, and not to some other value. But you haven't proven that the expression converges in the first place (and it is easy to see that it does not).


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