Real numbers arbitrary closeness (iterated fraction)

Question

I was playing with some iterated functions/fractions when I stumbled across this one $$\frac{1+\frac{1+\frac{1+...}{1-...}}{1-\frac{1+...}{1-...}}}{1-\frac{1+\frac{1+...}{1-...}}{1-\frac{1+...}{1-...}}}$$ which I found to be equal $i$. Am I correct? And if so, does it mean real numbers are not closed under arbitrary basic arithmetic (assuming there is no zero-division somewhere in it)?

Answer 1

As any field, the real numbers are definately closed under basic arithmetic operations. Namely for any two reals $x$ and $y$, $x+y$, $x-y$, $x\cdot y$ and, given that $y\neq 0$, $\frac{x}{y}$, are real numbers.

The reals are also complete, and from this it follows that any sequence of reals that converges in the complex numbers must converge to a real number. So I don't see how $$ a_{n+1}=\frac{1+a_n}{1-a_n} $$ can converge to $i$.