What is $\lim\limits_{k\to\infty}\frac{a_{k+1}}{a_{k}}$, where $a_k$ is the $k$th "generalized" fibonacci number? The following sequence is given:
\begin{align*}
a_1,a_2,..., a_n=1\\
a_{m}=\sum_{k=1}^{n}a_{m-k},m>n
\end{align*}
What is $\lim\limits_{k\to\infty}\frac{a_{k+1}}{a_{k}}$? Or how fast is the value of this expression growing with respect to $n$?
E.g. for $n=2$ it is $\frac{1+\sqrt{5}}{2}$, but is there a more general expression for any $n$?
Thank you
 A: The $n$-fibonacci numbers satisfy the recurrence
$$a_{m+n}=a_{m+n-1}+a_{m+n-2}+\cdots+a_{m+1}+a_{m}.$$
The algorithm listed here shows that, for some constants $c_1,\cdots,c_n$,
$$a_m=\sum_{i=1}^n c_i \alpha_i^m,$$
where $\alpha_1,\cdots,\alpha_n$ are the roots of the polynomial $x^n-\left(x^{n-1}+\cdots+x+1\right)$. You seek to find the root of this polynomial with the largest magnitude. 
This does not have a "nice" closed form as far as I am aware (although there might be some solution involving trigonometric functions; I'll edit this post if I come up with anything), but there is a way to determine the limit of this quantity as $n\to\infty$ (we call this $c$):
Multiplying this polynomial by $x-1$ gives the substantially easier to work with
$$x^{n+1}-2x^n+1.$$
It is clear that for all roots $z$ of this, $|z|^{n+1}\leq 1+2|z|^n$; however, for any real $r>2$, there exists an integer $n$ such that this is not satisfied. Thus, we obtain that $c\leq 2$. In addition, the polynomial evaluated at $2$ is positive while evaluated at $2-\epsilon$ it gives
$$1-\epsilon(2-\epsilon)^n,$$
which becomes negative for sufficiently large $n$; so for any real $r<2$ there exists a root between $r$ and $2$ if $n$ is large enough, so $c\geq 2$. We infer that $\boxed{c=2}$.
A: You can recover the generalized golden ratio here by expanding the limit:
$$\phi=\lim_{k\rightarrow\infty}\frac{F_{k+1}}{F_k}=\lim_{k\rightarrow\infty} \sum_{r=1}^{n}\frac{F_{k+1-r}}{F_k}=\lim_{k\rightarrow\infty} \sum_{r=1}^{n}\frac{F_{k+1-r}}{F_{k+1-r+1}}\frac{F_{k+1-r+1}}{F_{k+1-r+2}}\cdots \frac{F_{k-1}}{F_k}=1+\phi^{-1}+\phi^{-2}+\cdots+\phi^{-(n-1)}=\frac{1-(1/\phi)^{n}}{1-1/\phi}$$
This isn't explicitely solvable, but you'll find more discussion on this here: https://en.wikipedia.org/wiki/Generalizations_of_Fibonacci_numbers#Higher_orders
A: The generalized golden ratio is a root of $$x^n-x^{n-1}-...-x-1=0$$Which could be changed to $$x^{n+1}-2x^n+1=0$$Like in @CarlSchildkraut's answer. This could be obtained using the partial sum formula for geometric series. Anyways, divide by $x^n$: $$x+\frac{1}{x^n}=2$$As $n$ gets larger and larger, the second fraction approaches zero, so the largest $x$ must approach $2$.
Note that $x=1$ is a viable solution to the above equation.
