Finding the Limit of the Ratio of a Recursive Sequence's Terms Let {$f_n$} be defined recursively as $f_1 = f_2 = f_3 = 1$ and $f_n = f_{n-1} + f_{n-3}$ for all $n \gt 3$.
Also, define {$a_n$} as the ratio of the terms of {$f_n$}. That is, $a_n = \frac{f_{n+1}}{f_n}$ for some $n \geq 1$.
So, the terms of {$f_n$} are $$f_1 = 1,f_2 = 1,f_3 = 1,f_4 = 2,f_5 = 3,f_6 = 4,f_7 = 6,\ldots,$$ and the terms of {$a_n$} are $$a_1 = 1,a_2 = 1,a_3 = 2,a_4 = \frac{3}{2},a_5 = \frac{4}{3},a_6 = \frac{6}{4},\ldots$$
The question then becomes evaluating the limiting ratio of {$f_n$} or, in other words, 

Find $$\lim_{n \to \infty}{a_n} = \lim_{n \to \infty}\frac{f_{n+1}}{f_n}, \forall n \geq 1.$$

The way I approached this problem was to try to put bounds on $a_k = \frac{f_{k+1}}{f_k}$ for some $k$. It made the most sense to me that $1 \leq a_k \leq 2$ just based off of the first few terms of {$a_n$}. 
Then, I tried to rewrite $a_k = \frac{f_{k+1}}{f_k}$ in some way that would allow me to put bounds on $a_{k+1}$, since we want to show next that $1 \leq a_{k+1} \leq 2$.
$$a_{k+1} = \frac{f_{k+2}}{f_{k+1}} = \frac{f_{k+1} + f_{k-1}}{f_{k+1}} = 1 + \frac{f_{k-1}}{f_{k+1}}.$$
Next, I thought it would be a good idea to invert the inequality $1 \leq a_k \leq 2$. That is, $1 \geq \frac{1}{a_k} = \frac{f_k}{f_{k+1}} \geq \frac{1}{2}$ and then add $1$ to get the inequality $2\geq 1 + \frac{f_k}{f_{k+1}} \geq \frac{3}{2}$.
And while $1 + \frac{f_k}{f_{k+1}}$ looks like a pretty result, what I actually need to find in this case is $1 + \frac{f_{k-1}}{f_{k+1}}$.
It seems that at this point more clever manipulation is required, but I don't know what else can be done once I've reached this dead end. Can someone please elaborate on how to proceed with the above method or provide an alternative approach altogether?
I appreciate any and all advice!
Thanks for reading,
A
 A: Define the function $$G(x)=\sum_{n=1}^\infty f_n x^n$$
Through some standard generating function techniques, we get
$$G(x)={\frac {x}{1-x-x^3}}$$
where the series converges. The series has a radius of convergence equal to 
$\lim\limits_{n\to\infty}\frac{f_{n}}{f_{n+1}}$. However, the radius of convergence is also the distance to the closest singularity from the centre (in this case 0) in the complex plane. The second form of the function has singularities when the denominator is 0, i.e. $x^3+x-1=0$. The smallest in magnitude is the sole real root $r$ of the cubic. Equating these expressions for the radius of convergence, we have
$$\,\,\,\,\,\,\,\,\,\,\,\,\lim_{n\to\infty}\frac{f_{n}}{f_{n+1}}=r\\
\implies \lim_{n\to\infty}\frac{f_{n+1}}{f_{n}}=\frac1r$$
where $r\approx 0.68234$. Notably, this method also implicitly shows that the limit exists in the first place.
Alternative:
Somewhat more along your lines of thinking, let $L=\lim\limits_{n\to\infty}\frac{f_{n+1}}{f_n}$. Now
$$\large{\begin{align} 
\frac{f_{n+1}}{f_n}
&=\frac{f_{n}+f_{n-2}}{f_n}\\
&=1+\frac{1}{\frac{f_{n}}{f_{n-2}}}\\
&=1+\frac{1}{\frac{f_{n-1}+f_{n-3}}{f_{n-2}}}\\
&=1+\frac{1}{\frac{f_{n-1}}{f_{n-2}}+\frac{f_{n-3}}{f_{n-2}}}\\
&=1+\frac{1}{\frac{f_{n-1}}{f_{n-2}}+\frac{1}{\frac{f_{n-2}}{f_{n-3}}}}
\end{align}}$$ 
Notice that for very large $n$,$\frac{f_{n+1}}{f_n}\approx\frac{f_{n-k}}{f_{n-k-1}}$ for integer $k$. Hence we essentially have
$$L=1+\frac{1}{L+\frac{1}{L}}\\
\implies L^3-L^2-1=0$$
Since $L$ must be real and the above equation has one real root, the value of the limit is equal to the real root of the above polynomial.
A: You can find a closed-form for the sequence $f_n$ in terms of the roots ($z_1,z_2,z_3$) to the cubic equation $z^3-z^2-1=0$.  Two of these roots form a complex conjugate pair.  
The solution can be written as 
$$f_n=Az_1^n+Bz_2^n+Cz_3^n$$
where the constants $A$, $B$, and $C$ are found from the initial conditions on $f_1$, $f_2$, and $f_3$.
You should be able to directly find the coveted limit 
$$\lim_{n\to \infty}\frac{f_{n+1}}{f_n}=\lim_{n\to \infty}\left(\frac{Az_1^{n+1}+Bz_2^{n+1}+Cz_3^{n+1}}{Az_1^n+Bz_2^n+Cz_3^n}\right)$$
Hint:  The real root is positive, while the real part of the other roots is negative.
A: You have a Linear homogeneous recurrence relations with constant coefficients.  The standard approach is to assume a solution of the form $f_n=dr^n$ (traditionally the $d$ is an $a$, but I didn't want to use $a$ as you have already).  Now plug this into your recurrence, getting $dr^n=dr^{n-1}+dr^{n-3$}$ or $r^3-r^2-1=0$, which we call the characteristic polynomial.  You will have one solution for each root of the polynomial.  As the sum and multiple of solutions is again a solution, you have (here) a three dimensional vector space of solutions.  The one with the largest real part will dominate the long term behavior.  In this case you have one real root $\approx 1.4656$ and two conjugate complex roots $\approx -0.23279 \pm 0.79255i$.  As $n \to \infty$ the ratio of terms $a_n$ will approach $1.4656$ unless the initial conditions are carefully chosen to give that solution a zero multiplier.
