Define the $k$-bonacci numbers as such:
$$f^k_0=0, f^k_1=1, ..., f^k_{k-1}=k-1, k>1$$ $$f^k_{n+1}=\sum_{j=0}^{k-1} f^k_{n-j}, n\ge k$$
Then define $$r^k_n=\frac{f_{n+1}}{f_n}$$
Then if $L_k=\lim_{n\rightarrow \infty}r^k_n$, what is $L_k$?
I am assuming that the limit exists. My question is the process. If $k=2$, then we have
$$L_2=\lim_{n\rightarrow \infty}r^2_n=\lim_{n\rightarrow \infty}\frac{f^2_{n+1}}{f^2_n}=\lim_{n\rightarrow \infty}\frac{f^2_{n}+f^2_{n-1}}{f^2_{n}}=1+\frac{1}{\lim_{n\rightarrow \infty}\frac{f^2_{n}}{f^2_{n-1}}}=1+\frac{1}{L_2}$$
and then you can use the quadratic formula to solve. You can do the same thing for $k=3$, albeit much more work...
$$L_3=\lim_{n\rightarrow \infty}\frac{f^3_{n+1}}{f^3_n}=\lim_{n\rightarrow \infty}\frac{f^3_{n}+f^3_{n-1}+f^3_{n-2}}{f^2_{n}}=1+\frac{1}{L_3}+\frac{1}{\lim_{n\rightarrow \infty}\frac{f^3_{n}}{f^3_{n-2}}}$$
$$=1+\frac{1}{L_3}+\frac{1}{\lim_{n\rightarrow \infty}\frac{f^3_{n-1}+f^3_{n-2}+f^3_{n-3}}{f^3_{n-2}}}=1+\frac{1}{L_3}+\frac{1}{L_3+1+\lim_{n\rightarrow \infty}\frac{f^3_{n-3}}{f^3_{n-2}}}$$
$$=1+\frac{1}{L_3}+\frac{1}{L_3+1+\frac{1}{\lim_{n\rightarrow \infty}\frac{f^3_{n-2}}{f^3_{n-3}}}}=1+\frac{1}{L_3}+\frac{1}{L_3+1+\frac{1}{L_3}}$$
Rearranging then requires solving the quartic $x^4-2x^2-2x-1=0$, which is not too bad. However, this method seems to fall apart. Is it possible to prove (is it even true?) that this process does not work for $k>3$? Or does it work and it just is a relatively large sum of continued fractions?