What's the value of $n+\frac{n}{n+\frac{n}{n+\frac{n}{\ddots}}}$ for $n\in\mathbb{C}$? Write $$\phi_n\stackrel{(1)}{=}n+\cfrac{n}{n+\cfrac{n}{\ddots}}$$ so that $\phi_n=n+\frac{n}{\phi_n},$ which gives $\phi_n=\frac{n\pm\sqrt{n^2+4n}}{2}.$ We know $\phi_1=\phi$, the Golden Ratio, so let's take $\phi_n\stackrel{(2)}{=}\frac{n+\sqrt{n(n+4)}}{2}$. (Is that justified?)
Wolfram Alpha states that, with $(2)$, $$\lim\limits_{n\to -\infty}\phi_n=-1.$$ Why? Can I infer that this is true for $(1)$ and, if so, why?
I wonder what happens in $(1)$ for $n\in\mathbb{C}\backslash\mathbb{Z}$ too. I got something horrendous looking in $(2)$ for $n=i$.

Clarification: I'm trying to find $\phi_n$ in terms of $n$. See the comments below.
 A: For any given $n \ne 0, -4$, let $( a_{n,m} )_{m\in\mathbb{Z}_{+}}$ be the sequence
defined by
$$a_{n,m} = \begin{cases}n,& m = 1\\\displaystyle n + \frac{n}{a_{n,m-1}},&m > 1\end{cases}$$
Let $\displaystyle\;\mu_n = \frac{n+\sqrt{n(n+4)}}{2}\;$ and
$\displaystyle\;\nu_n = \frac{n-\sqrt{n(n+4)}}{2}\;$. It is easy to verify the following expression
$$a_{n,m} = \frac{\mu_n^{m+1} - \nu_n^{m+1}}{\mu_n^{m}-\nu_n^{m}}$$
provided a closed-form solution for $a_{n,m}$.
On those portion of complex plane where $|\mu_n|$ differs from $|\nu_n|$,
if is clear one of $\mu_n$ or $\nu_n$ will completely dominate the other one
for large $m$. As a result, we can make the following partial summary about $\phi_n$.
$$\phi_n = \lim_{m\to\infty} a_{n,m} = 
\begin{cases}
\mu_n, & |\mu_n| > |\nu_n|\\
\\
???& |\mu_n| = |\nu_n|\quad\leftarrow
\begin{array}{c}
\small\verb/This includes the special/\\
\small\verb/special case when /n = 0, -4.
\end{array}
\\
\nu_n, & |\mu_n| < |\nu_n|
\end{cases}
$$
In particular,
$$
\begin{array}{lcl}
n \in (0,\infty)   & \implies & |\mu_n| > |\nu_n| \implies \phi_n = \mu_n\\
n \in (-\infty,-4) & \implies & |\mu_n| < |\nu_n| \implies \phi_n = \nu_n
\end{array}
$$
and hence
$$\lim\limits_{n\to-\infty}\phi_n = \lim\limits_{n\to-\infty}\nu_n
= \lim\limits_{k\to\infty}\frac{-k+\sqrt{k(k-4)}}{2}
= \lim\limits_{k\to\infty}\frac{-2k}{k+\sqrt{k(k-4)}} = -1
$$
Update
Let us switch to the case $|\mu_n| = |\nu_n|$ but $n \ne 0, -4$. In particular, this cover
the range where $n \in (-4,0)$. Since $\mu_n\nu_n = -n$ and $\mu_n \ne \nu_n$, we can find
a $\theta_n \in (0,\pi)$ such that
$$\big\{\; \mu_n, \nu_n \;\big\} = \big\{\; \sqrt{-n}e^{i\theta_n}, \sqrt{-n}e^{-i\theta_n}\;\big\}$$
In terms of $\theta_n$, we have
$$a_{n,m} = \sqrt{-n}\frac{\sin((m+1)\theta_n)}{\sin(m\theta_n)}$$
There are two sub-cases. 


*

*If $\displaystyle\;\frac{\theta_n}{\pi} \in \mathbb{Q}$, then $a_{n,m}$ is periodic in $m$. Notice when $\theta \in (0,\pi)$, $\displaystyle\;\frac{\sin((m+1)\theta)}{\sin\theta}$ is never an constant sequence. So $\phi_n$ diverges.

*If $\displaystyle\;\frac{\theta_n}{\pi} \notin \mathbb{Q}$, then $a_{n,m}$ form a dense subset of the line $\big\{\; \sqrt{-n} t : t \in \mathbb{R} \;\big\} \subset \mathbb{C}$.
Once again, $\phi_n$ diverges.


Finally, it leaves us the cases $n = 0$ and $n = -4$. 


*

*For $n = 0$, it is easy because start at $m = 2$, we encounter an undefined expression
like $a_{0,2} = 0 + \frac{0}{0}$. So all $a_{0,m}, m \ge 2$ and hence $\phi_0$ are undefined.

*For $n = -4$, we have $\mu_{-4} = \nu_{-4} = -2$. Notice
$$\lim_{\mu\to\nu}\frac{\mu^{m+1}-\nu^{m+1}}{\mu^{m}-\nu^{m}} = \mu\frac{m+1}{m}$$
We will suspect $\displaystyle\;a_{-4,m} = -2\frac{m+1}{m}$. By direct substitution,
one can check that this is indeed the case. As a result,
$$\phi_{-4} = \lim\limits_{m\to\infty} a_{-4,m} = -2\lim_{m\to\infty}\frac{m+1}{m} = -2 = \nu_n$$
Combine all these, we obtain:
$$\phi_n = \lim_{m\to\infty} a_{n,m} = 
\begin{cases}
\mu_n, & |\mu_n| > |\nu_n|\\
\nu_n, & |\mu_n| < |\nu_n|\quad\text{ or }\quad n = -4\\
\text{undefined},& n = 0\\
\text{diverges},& |\mu_n| = |\nu_n|
\end{cases}
$$
Update2
The final question is what are the set that $|\mu_n| = |\nu_n|$. 
It turns out when $n \ne 0$,
$$\begin{align}
|\mu_n| = |\nu_n|
&\iff  \left|1 + \sqrt{1+\frac{4}{n}}\right| = \left|1 - \sqrt{1+\frac{4}{n}}\right|
\iff \Re\left(\sqrt{1+\frac{4}{n}}\right) = 0\\
&\iff
n \in [-4,0)
\end{align}$$
In fact, if we define $\lambda(z)$ by
$$\mathbb{C}\setminus (-4,0] \ni z\quad\mapsto\quad \lambda(z) = \frac{z}{2}\left(1 + \sqrt{1 + \frac{4}{z}}\right) \in \mathbb{C},$$
$\lambda(z)$ will be a single-valued function over $\mathbb{C} \setminus (-4,0]$ and analytic over its interior $\mathbb{C} \setminus [-4,0]$.
Furthermore, $\lambda(n)$ coincides with $\mu_n$ and $\nu_n$ on $(0,\infty)$ and $(\infty,-4]$ respectively! What this means is the apparent switching of value of $\phi_n$ 
between $\mu_n$ and $\nu_n$ is really an artifact of how we label them.
At the end, we have a much simpler description for $\phi_n$.
$$\phi_n = \begin{cases} 
\frac{n}{2}\left(1 + \sqrt{1 + \frac{4}{n}}\right),& n \notin (-4,0]\\
\\
\text{ undefined/diverges },& n \in (-4,0]
\end{cases}$$
A: Let $f_x(t) = x+\frac{x}{t}$ (I'm using $x$ rather than $n$ since we're interested in continuous behavior now, and $f_n$ is a little confusing as a family of functions rather than a sequence); the continued fraction thus corresponds to the sequence $\{t_0 = x, t_n = f_x(t_{n-1})\}$.  Then $\frac{df}{dt} = -\frac{x}{t^2}$; since the absolute value of this is larger than $1$ in a neighborhood of $t=-1$ once $x$ gets large — and in fact will be larger than $1$ around any finite value of $t$ once $x$ is large enough — then no fixed point of $f_x$ that's bounded as $x\to-\infty$ can be stable.  Since $\{t_n\}$, if it converges, must converged to a fixed point of $f_x$, this implies that the sequence can't converge as $x\to -\infty$.
A: Found why it's $-1$.
Rewrite the equation as :$$n\bigg(\frac{1+\frac1n\sqrt{n^2 + 4n}}{2}\bigg)$$
The square root can be written as $\sqrt{n^2(1+4/n)} = |n|\sqrt{1+4/n} = -n\sqrt{1+4/n}$ because $n<0$.
Then you obtain :
\begin{eqnarray*}
\phi(n) &=& \frac{n}{2}\big(1-\sqrt{1+4/n}\big)\\
&=& \frac n2(1-(1+\frac2n) + O(n^{-2}))\\
&\to&-1
\end{eqnarray*}
A: If $f_n(x) = n+n/x$, I'm not sure about your notation, but I think $(1)$ defines $\phi_n$ as the limit (provided it exists) of $f_n \circ f_n \circ \ldots \circ f_n (n)$.
Such a limit would be a fixpoint of $f_n$, which means one of the two complex numbers $y_n,z_n = \frac {n \pm \sqrt {n(n+4)}}2$.
To guess which one it is, we should wonder which one is the attractive fixpoint ,and which one is repulsive.
$f_n'(x) = -n/x^2 = y_nz_n/x^2$, and so $|f'_n(y_n)| = |z_n/y_n| = 1/|f'_n(z_n)|$. To be attractive, the derivative at that point has to be small ($< 1$ in modulus), hence the solution with the bigger modulus will be attractive, and that's the one the sequence will converge to (unless the sequence accidentally ends up exactly on the smaller fixpoint).
One can check that the two fixpoints have the same modulus when $n \in [-4;0]$, so we can't really define what's $\phi_n$ for those values of $n$. Just look at the sequence for such an $n$, you will see it has chaotic behaviour. On the rest of the complex plane, by picking the bigger fixpoint you obtain a continuous function $n \mapsto \phi_n$ defined on $\Bbb C \setminus [-4;0]$.
So $(2)$ is only correct for $n>0$, and you have to switch the sign of the square root for $n < -4$
A: The usual trick is to apply the third binomial formula, so that
$$
\frac{n+\sqrt{n(n+4)}}{2}
=\frac{n^2-(n^2+4n)}{2(n-\sqrt{n(n+4)})}
=-\frac{2 |n| }{|n|+\sqrt{|n|(|n|-4)}}
$$
Now standard limit procedurs for fractions apply, cancel $|n|$ in numerator and denominator, move the limit inside the square root in the denominator, ..., to get the limit $-1$.

To the general question, I do not think that the convergence of this continued fraction is well-defined. Depending on the order of the convergent, one can cancel the $n$ to either get
$$
n+\frac{n}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\vdots+1}}}}
$$
or
$$
n+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\vdots+1}}}},
$$
so that these even and odd subsequences of convergents converge to $n\cdot\phi_1$ and $n-1+\phi_1$. 

The convergents in the tradional sense form, for $n\ne 1$, an oscillating sequence.

Cancellation leaves some more of the $n$ in place
$$
n+\frac{n}{n+\frac{1}{1+\frac{1}{n+\frac{1}{\vdots+1}}}}
$$
or
$$
n+\frac{1}{1+\frac{1}{n+\frac{1}{1+\frac{1}{\vdots+1}}}},
$$
so that after cancellation one obtains a classical continued fraction $[n;1,n,1,n,...]$ or  $n\cdot[1;n,1,n,1,...]$, which do indeed converge for integer $n\ge 1$ in the classical sense.
The resulting recursion
$$
a_{k+1}=F(a_k)=n+\frac1{1+\frac1{a_k}}
$$
is contractive on $a_k\in (-\infty,-2)\cup(0,\infty)$. For $n\in\Bbb R\setminus[-4,0]$ the fixed point
$$
a_*=\frac n2\left(1+\sqrt{1+\frac 4n}\right)
$$
falls inside this contraction region and is thus the  value of the given continued fraction.
