Confusion while evaluating the complex continued fraction $\frac{1}{i+\frac{1}\ddots}$

Well, Vasily has put his finger on one problem, but I would like to point out a much more serious one.

We write a continued fraction to get a number that is the limit of the convergents, that is, of the expressions that you get when you cut your continued fraction off, to be a finite c.f.

The first convergent is $\frac1i$, no problem, but the second is $$\frac1{i+\frac1i}=\frac10\,,$$ a most unfortunate development. My recommendation would be to go on to other complex continued fractions, where the partial denominators are rather larger than $i$.

The problem with this expression is that function $f(x)=(i+x)^{-1}$ is a tricky one:

$$ f(f(x)) = \frac{1}{i+\frac{1}{i+x}}=\frac{i+x}{-1+ix+1}=-i+\frac1x,\\ f(f(f(x))) = \frac{1}{i+(-i+1/x)}=x. $$

So every number generates the orbit of length 3, except for two numbers you have found, which are fixed points. Thus, the series of nesting functions $f$ does not converge if you haven't already started with a fixed point. So the question as it is stated has no sense: you cannot assign any number to this expression.

I shall answer the general question completely. The continued fraction $[x;x,x,...]$ converges for any $x ∈ \mathbb{C} ∖ i(-2,2)$ $= \{ z : z ∈ \mathbb{C} ∧ z ∉ \{ ir : r ∈ (-2,2) \} \}$, and here is a proof sketch.

If $[x;x,x,...]$ converges to $c$, then $c = x + 1/c$ and hence $c$ is a root of the quadratic $( t ↦ t^2 - x t - 1 )$.

Let $r,s$ be the roots of the quadratic $( t ↦ t^2 - x t - 1 )$ and so $r + s = x$ and $r s = -1$.

Let the sequence of approximants be $(a_n)$ where $a_1 = x$ (and the sequence stops if it becomes $0$).

Let $b_0 = 1$ and $b_n = a_n b_{n-1}$ for each $n$ such that $a_n$ is defined, giving $a_n = \frac{b_n}{b_{n-1}}$.

Given any $n ∈ \mathbb{N}^+$ such that $a_n ≠ 0$:

  $b_{n+1} = ( x + \frac1{a_n} ) b_n = x b_n + b_{n-1}$.

  Thus $b_{n+1} - r b_n = s ( b_n - r b_{n-1} ) = s^n ( b_1 - r b_0 ) = (x-r) s^n = s^{n+1}$.

  Thus $b_n - r^n b_0 = \sum_{k=1}^n r^{n-k} s^k$ and hence $b_n = \sum_{k=0}^n r^{n-k} s^k$.

  If $r ≠ s$, then $b_n = {\large \frac{ r^{n+1} - s^{n+1} }{r-s} }$ and hence $a_n = {\large \frac{ r^{n+1} - s^{n+1} }{ r^n - s^n } }$.

  If $r = s$, then $b_n = (n+1) r^n$ and hence $a_n = \frac{n+1}{n} r$.

If $x ∉ i[-2,2]$,

  $|r| ≠ 1$ otherwise $x = r + s = r - \frac{1}{r} = r - r^* = 2i·Im(r) ∈ i[-2,2]$.

  Permute $r$,$s$ such that $|r| > 1 > |s|$, possible since $|r| · |s| = |rs| = 1$.

  Then by induction we get $a_n = {\large \frac{ r^{n+1} - s^{n+1} }{ r^n - s^n } } ≠ 0$ for every $n ∈ \mathbb{N}^+$.

  Thus $a_n = r + {\large \frac{ (r-s) s^n } { r^n - s^n } } = r + {\large \frac{r-s}{ (\frac{r}{s})^n - 1 } } \to r$ as $n \overset{∈\mathbb{N}}\to \infty$.

If $x ∈ \{2i,-2i\}$,

  $r = s ∈ \{i,-i\}$ because $(r-s)^2 = (r+s)^2 - 4rs = 0$.

  Then by induction we get $a_n = \frac{n+1}{n} r ≠ 0$ for every $n ∈ \mathbb{N}^+$.

  Thus $a_n \to r$ as $n \overset{∈\mathbb{N}}\to \infty$.

If $x ∈ i(-2,2)$,

  By induction we get $a_n ∈ i\mathbb{R}$ whenever $a_n$ is defined.

  If $a_n \to c$ as $n \to \infty$:

    $c ∈ i\mathbb{R}$ since $i\mathbb{R}$ is closed.

    But $c ∈ \{r,s\} = {\large \frac{ x \pm \sqrt{x^2+4} }{2} }$ and hence $c ∉ i\mathbb{R}$ since $x^2+4 > 0$.

    Contradiction.

  Therefore $( a_n )$ either terminates in a $0$ or does not converge.

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It is worth emphasizing that one cannot assume that the continued fraction converges. If it converges then its limit must be one of the roots of the quadratic (and the above proof shows us explicitly which one). But it may be that it does not converge in the first place, such as for $x = i$.