As I'm sure many of you know, the golden ratio is $\frac{1+\sqrt{5}}{2}$, the positive root of the polynomial $x^2-x-1$. Similarly, the positive root of $x^3-x^2-x-1$ is the tribonacci number $\approx 1.839$. In general the positive root of the polynomial $x^n-\sum_{k=0}^{n-1} x^k$ is the ratio $f_{m+1}/f_{m}$ to which the sequence $f_m = \sum_{k=1}^{n} f_{m-k}$ converges.
As we increase $n$ in the polynomial $x^n-\sum_{k=0}^{n-1} x^k$, the positive root approaches $2$. I am interested in finding the rate at which this happens.
We can actually considerably simplify this problem by considering that $$\sum_{k=0}^{n-1} x^k = \frac{x^n-1}{x-1}$$ The question then reduces to solving $x^n-\frac{x^n-1}{x-1} = 0$. We can get $$x^{n+1}-x^n-(x^n-1) = x^{n+1}-2x^n+1=0$$
The question finally reduces to finding the asymptotic of the positive root of $x^{n+1}-2x^n+1$ (let's call this function $g(n)$).
Any ideas?
Edit: I graphed $\frac{1}{2-g(n)}$ for $n = 1...30$ in python using matplotlib. This looked exponential, so I took the log of it. The log then seemed to converge to $\ln(2)$, giving me an asymptotic of $g(n)=2-2^{-n}$. If anyone could give me more error terms or explain why this is, feel free to do so.
Let $h_n(x)=x^{n+1}-2x^n+1$. Note that $h_n(2)=1$ and $h_n'(2)=2^n$. This immediately gives $2-2^{-n}$ as an approximation to the root $g(n)$, by taking a linear approximation to $h_n$ near $2$.
To get bounds on how good this approximation is, note first that $$h_n'(x)=((n+1)x-2n)x^{n-1}$$ is less than $2^n$ for $x$ strictly between $2-2^{-n}$ and $2$ and so actually $g(n)<2-2^n$. To get a lower bound on $g(n)$, notice that $$h_n'(2-a)=(2-(n+1)a)(2-a)^{n-1}>2^n-na2^n$$ for $0<a<\frac{2}{n+1}$ (here we use the fact that $\prod_{i=1}^n(x-a_i)>x^n-x^{n-1}\sum_{i=1}^n a_i$ if $0<a_i<x$ for all $i$, which is easily proved by induction on $n$). In particular, for any fixed $c>1$, then for $n$ sufficiently large, $h_n'(2-a)>2^n/c$ as long as $0<a<c2^{-n}$. It follows that $h_n(2-c2^{-n})<0$ and so $g(n)>2-c2^{-n}$.
That is, for any $c>1$, we have $$2-c2^{-n}<g(n)<2-2^{-n}$$ for all sufficiently large $n$. In particular, $\frac{2-g(n)}{2^{-n}}\to 1$ as $n\to\infty$.