In any general recurrence of the form:
$$x_n = \sum_{i=1}^k {a_i x_{n-i}}$$
you can determine an explicit formula in terms of the roots of the polynomial of degree k:
$$z^k - a_1{z^{k-1}} - a_2{z^{k-2}} - a_{k-1}z - a_k$$
If the polynomial has no repeated roots, then the form of the explicit formula will be:
$$x_n = b_1 r_1^n + b_2 r_2^n + ... + b_k r_k^n$$
Where the $b_i$ can be any numbers, and the $r_i$ are the distinct roots of the polynomial.
This means that "most of the time," $\lim\limits_{n\to \infty}{x_{n+1}/x_n}$ will be equal to the root $r_i$ with the largest absolute value such that $b_i\neq 0$. If two $r_i$ have the same largest absolute value, the convergent behavior might be odd - it possibly might never converge.
There is a lot of linear algebra involved in this - the $r_i$s are eigenvalues of a matrix which sends $(x_i,x_{i+1},...,x_{i+k-1})$ to $(x_{i+1},...,x_{i+k})$
In the case of what you call the k-nacci numbers, your polynomial is:
$$x^k-x^{k-1}-x^{k-2}-..-1 = x^k -\frac{x^k-1}{x-1}$$
I'm not sure what you can say about the roots of this polynomial when $k>2$. It's easy to show it has no repeated roots (a polynomial has no repeated roots of it is relatively prime to its derivative.)
The second term vanishes quite quickly. We have $F_n=\lfloor \frac{\phi^n}{\sqrt{5}}+\frac{1}{2}\rfloor$, where $\lfloor \cdot \rfloor$ denotes the floor function. If the floor weren't there, you could solve for $n$, and if you rounded you would be wrong only for possibly very small $n$.
Update: A quick calculation shows that $n=\left[ \log_\phi \sqrt{5}(F_n-\frac{1}{2}) \right]$ holds for all $n\ge 3$, and fails for $n=1,2$. In this case $[\cdot]$ denotes the rounding function, or $[x]=\lfloor x+\frac{1}{2}\rfloor$.
Best Answer
The Fibonacci numbers are defined by a second-order linear recurrence equation: $$F_{n+2} = F_{n+1} + F_n$$ This means we can treat the solution of $F_n$ in terms of $n$ as a problem in linear algebra involving only $2$-dimensional vectors. In some sense, what we are doing is modelling this as a dynamical process on a $2$-dimensional state space.
Let $V = \mathbb{R}^2$. We define a linear operator $T : V \to V$ by $$T(x, y) = (y, x + y)$$ Notice that $T(F_{n}, F_{n+1}) = (F_{n+1}, F_{n+2})$, so you can think of $V$ as being a sliding $2$-entry window on the Fibonacci sequence and $T$ as the operator which advances the window along the Fibonacci sequence. The initial conditions $F_0 = 0, F_1 = 1$ then imply that $$T^n(0, 1) = (F_n, F_{n+1})$$ so all we need to do to find $F_n$ in terms of $n$ is to find an effective way to compute iterates of the operator $T$!
Now, we get our hands dirty and represent $T$ as a matrix: $$T = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$$ Imagine if we could somehow find an invertible matrix $P$ and a diagonal matrix $D$ such that $T = P D P^{-1}$; then, by matrix algebra, we would have $T^n = P D^n P^{-1}$, and it is easy to compute powers of diagonal matrices. The theory of eigenvectors and eigenvalues gives us one way to find such a factorisation of $T$. Notice that $$\det (T - x I) = \det (P(D - x I)P^{-1}) = (\det P)(\det (D - x I))(\det P^{-1}) = \det (D - x I)$$ but a simple calculation shows $$\det (D - x I) = \det \begin{pmatrix} \lambda_1 - x & 0 \\ 0 & \lambda_2 - x \end{pmatrix} = (\lambda_1 - x)(\lambda_2 - x)$$ so whatever $\lambda_1$ and $\lambda_2$ are, they must be the zeros of the polynomial $$\det (T - x I) = \det \begin{pmatrix} -x & 1 \\ 1 & 1 - x \end{pmatrix} = x^2 - x - 1$$ which, surprise surprise, is the minimal polynomial of the golden ratio. So let $\lambda_1 = \frac{1}{2} (1 + \sqrt{5})$ and $\lambda_2 = \frac{1}{2} (1 - \sqrt{5})$. These are the eigenvalues of $T$. By construction, $\det (T - \lambda_1 I) = \det (T - \lambda_2 I) = 0$, so there must be non-zero vectors $\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ and $\begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$ such that $$ T \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \lambda_1 \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} $$ $$ T \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \lambda_2 \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} $$ These vectors are called the eigenvectors of $T$. I leave you to verify that $x_1 = \lambda_1 - 1$, $y_1 = 1$, $x_2 = \lambda_2 - 1$, $y_2 = 1$ works, but there are other solutions.
Define the matrix $P$ by $$P = \begin{pmatrix} x_1 & x_2 \\ y_1 & y_2 \end{pmatrix}$$ Notice that $$\begin{pmatrix} x \\ y \end{pmatrix} = \alpha_1 \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + \alpha_2 \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = P \begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix}$$ so by linearity we have $$T \begin{pmatrix} x \\ y \end{pmatrix} = \lambda_1 \alpha_1 \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + \lambda_2 \alpha_2 \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = P \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix}$$ but we can invert $P$ to find $\alpha_1$ and $\alpha_2$ in terms of $x$ and $y$, so we have obtained the desired factorisation of $T$ as $P D P^{-1}$ with $D$ diagonal. Putting everything together, we get the formula $$T^n = P \begin{pmatrix} {\lambda_1}^n & 0 \\ 0 & {\lambda_2}^n \end{pmatrix} P^{-1}$$ and applying this to the vector $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ gives Binet's formula.