The n-nacci numbers are defined as $${}_nF_k = {}_nF_{k - 1} + {}_nF_{k - 2} + \cdots + {}_nF_{k - n + 1}$$
Now, it's pretty well-known that the limit of successive $2$-nacci numbers (i.e. the Fibonacci numbers) is equal to $$\lim_{n\to\infty} {F_{n+1} \over F{n}} = \phi = \frac{1 + \sqrt{5}}{2}$$.
How would one find the limiting ratio of the tribonacci numbers? And how can that be generalized to the n-nacci numbers?
On
The usual solution to recurrence relations works here. Find the characteristic polynomial and its greatest root(s) in absolute value. Here you have $a_k=a_{k-1}+a_{k-2}+\dots+a_{k-n}$ so the polynomial is $x^n=\frac {x^n-1}{x-1}$
On
Denote the limiting ratio of $n$-nacci numbers as $\,a_n.\,$ A standard result is that $\,a_n\,$ is a root of a polynomial equation. More precisely, define
$$ f_n(x) := x^{n} - \sum_{k=0}^{n-1} x^{k} = \frac{g_n(x)}{x-1} \tag1 $$
where
$$ g_n(x) := 1 - (2 -x)x^n \tag2 $$
by expanding $\,f_n(x)(x-1)\,$ as a telescoping sum simplifying to $\,g_n (x).\,$
The result is that $a_n$ is the largest real root of $\,f_n(x) = 0.\,$
Assuming that $\,x \ne 1\,$ then $\,f_n(x) = 0\,$ iff $\,g_n(x) = 0.\,$
Assuming that $\,x \ne 2\,$ then $\,g_n(x) = 0\,$ iff $\,x = 2-x^{-n}.\,$
Assuming that $\,n>1\,$ then $\,f_n(1) = 1 - n < 0\,$ and $\,f_n(2) = g_n(2) = 1 > 0.\,$
Thus, by the intermediate value theorem there exists $\,1 < a_n < 2\,$ such that $\,f_n(a_n) = 0.\,$
Now, since $\,a_n = 2 - a_n^{-n},\,$ this implies that $\,a_n\to 2$ but is there a better result? Yes.
Define
$$ b_n := \frac{2^{-n}+2-a_n}2 = h_n(2^{-n-1}) \tag3 $$
where
$$ h_n(z) := 2z + z\sum_{k=1}^\infty c_n(k) \frac{z^k}{k!} \tag4 $$
and
$$ c_n(k) := n \prod_{j=1}^{k-1} j+n+k\,n. \tag5 $$
You could be interested by the following link
http://mathworld.wolfram.com/Fibonaccin-StepNumber.html
What they essentially show is that the limit of the ratio is the solution of the equation $$x^n(2-x)=1$$ For tribonacci numbers, $x$ is given by $$\frac{1}{3} \left(1+\sqrt[3]{19-3 \sqrt{33}}+\sqrt[3]{19+3 \sqrt{33}}\right)=1.83929$$ For tetranacci numbers, we start with the ugly formula,
$$x =\frac{1}{4}+\frac{1}{2}\sqrt{z}+\frac{1}{2}\sqrt{\frac{11}{4}-z+\frac{13}{4\sqrt{z}}}$$
where,
$$z = \tfrac{11}{12}-\tfrac{1}{3}\left(\tfrac{65+3\sqrt{1689}}{2}\right)^{1/3}+\tfrac{1}{3}\left(\tfrac{-65+3\sqrt{1689}}{2}\right)^{1/3}$$
and which is $x=1.92756\dots$. If $n$ increases, the limit of the ratio is $2$.
For high orders, the solution of the equation can be approximated using Newton method starting at $x=2$ and the first iterate is simply $(2-2^{-n})$ which is $1.9375$ for tetranacci numbers. A better approximation is $(2-2^{-n}-2^{-2 n-1} n)$ which gives $1.92969$.
Going to higher order of Newton method, we can approximate still better using $$x \simeq 2+\frac{1}{\frac{(n+1) n}{2^{n+3}-4 n}+\frac{n}{2}-2^n}$$ which gives $1.84049$ for tribonacci numbers and $1.92765$ for tetranacci numbers.
Continuing the work, I found that this limit could be accurately represented by $$x \simeq 2-\exp\left({0.0965617 -0.73092 n+\frac{0.80957}{n}}\right)$$
Edit six years later !
Back to the problem (almost by accident), I found a rather better approximation $$x\sim 2+\frac{1 }{\frac{n}{2}+\frac{\left(3\ 2^n-n+1\right) (n+1) n}{3 \left(n^2-\left(2^{n+3}+1\right) n+2^{2 n+3}\right)}-2^n } $$
$$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 2 & 1.6250000000000000000 & 1.6180339887498948482 \\ 3 & 1.8394879369768586903 & 1.8392867552141611326 \\ 4 & 1.9275687815833801235 & 1.9275619754829253043 \\ 5 & 1.9659485002996453378 & 1.9659482366454853372 \\ 6 & 1.9835828545045254017 & 1.9835828434243263304 \\ 7 & 1.9919641970897590225 & 1.9919641966050350211 \\ 8 & 1.9960311797568978058 & 1.9960311797354145898 \\ 9 & 1.9980294702632357105 & 1.9980294702622866987 \\ 10 & 1.9990186327101425508 & 1.9990186327101011387 \\ 11 & 1.9995104019782872690 & 1.9995104019782854914 \\ 12 & 1.9997555009373176116 & 1.9997555009373175367 \\ 13 & 1.9998778327115455431 & 1.9998778327115455400 \\ 14 & 1.9999389387495946486 & 1.9999389387495946486 \end{array} \right)$$ For $n \geq 14$, the first twenty figures are exact.