Let be the polynomial:
$P_n (x)=x^{n+1} - (x^{n-1}+x^{n-2}+\cdots+x+1)$
I want to prove that it has a single positive real root we'll denote by $x_n$, and then to compute: $\lim_{n\to\infty} x_{n}$
Let be the polynomial:
$P_n (x)=x^{n+1} - (x^{n-1}+x^{n-2}+\cdots+x+1)$
I want to prove that it has a single positive real root we'll denote by $x_n$, and then to compute: $\lim_{n\to\infty} x_{n}$
Since it's not much more work, let's study the roots in $\mathbb{C}$.
Note that $x=1$ is not a solution unless $n=1$, since $P_n(1) = 1-n$.
Since we are interested in the limit $n\to\infty$, we can assume $x\ne 1$. Sum the geometric series, $\begin{eqnarray*} P_n (x) &=& x^{n+1} - (x^{n-1}+x^{n-2}+\cdots+x+1) \\ &=& x^{n+1} - \frac{x^n-1}{x-1}. \end{eqnarray*}$ The roots will satisfy $x_n^{n}(x_n^2-x_n-1) = -1.$
(Addendum: If there are concerns about convergence of the sum, think of summing the series as a shorthand that reminds us that $(x-1)P_n(x) = x^{n}(x^2-x-1) + 1$ for all $x$.)
If $0\le |x_n|<1$, $\lim_{n\to\infty} x^n = 0$, thus, in the limit, there are no complex roots in the interior of the unit circle.
If $|x_n|>1$, $\lim_{n\to\infty} 1/x^n = 0$, thus, in the limit, the roots must satisfy $x_n^2 - x_n - 1 = 0.$ There is one solution to this quadratic equation with $|x_n|>1$, it is real and positive, $x_n = \frac{1}{2}(1+\sqrt{5}).$ This is the golden ratio. It is the only root exterior to the unit circle.
The rest of the roots must lie on the boundary of the unit circle.
|P_{15}(x+i y)|.">
Figure 1. Contour plot of $|P_{15}(x+i y)|$.
$P_n(x)=x^{n+1}-(x^{n-1}+x^{n-2}+...x+1)=x^{n+1}-\sum_{k=0}^{n-1}x^{k}$ By the formula for the sum of a finite geometric series, we have, if x $\ne$ 1: $P_n(x)=x^{n+1}-\frac{x^n-1}{x-1}$ Equating to zero, $x^{n+1}=\frac{x^n-1}{x-1} \implies x^{n+1}(x-1)=x^{n+2}-x^{n+1}=x^n-1 \implies x^{n+2}-x^{n+1}-x^{n}=-1$ So, provided x $\ne$ 0, we also have $x^2-x-1=\frac{-1}{x^n}$
Now we can limit both sides as n approaches infinity, to get (assuming x>1) $x^2-x-1=0 \implies x=\frac{1+\sqrt{5}}{2}$
Let us establish the uniqueness of this root for x>1: $P_n'(x)=(n+1)x^{n+1}-((n-1)x^{n-1}+(n-2)x^{n-2}+...2x+1)=0$ Has exactly one positive root, by Descartes' rule of Signs. Note that $P_n'(0)=1$ and $p_n'(1)=n+1-(n-1+n-2+n-3...)<0$ so the root must lie between 0 and 1. As a result, the curve intersects the positive x axis in exactly one place.
$P_n(\phi)=\frac{1}{\phi^n(\phi-1)}(\phi^{2}-\phi-1+\frac{1}{\phi^n})$ Now, note that $\phi^2-\phi-1=0$ by definition, and $\frac{1}{\phi^n}=(\phi-1)^n$, so $P_n(\phi)=\frac{(\phi-1)^{n-1}}{\phi^n}=(\phi-1)^{2n-1}$ Since, $|\phi-1|<1$, as n tends to infinity, $P_n(\phi)$ tends to zero.
I've never done any proper real analysis (I'm 17) so forgive me for not being rigorous. Is this sufficient?
Descartes' rule of signs shows that $P_n$ has at most one positive root, and since $P_n(1)=1-n<0$ and $\lim_{+\infty} P_n=+\infty$, $P_n$ has exactly one positive root $x_n>1$. Let $R_n(x)=x^{-n}(x-1)P_n(x)=x^2-x-1+x^{-n}$ Then $R_n$ has exactly two positive roots 1 and $x_n$, so that it is negative on the interval $(1,x_n)$ and positive on $(x_n,+\infty)$: for all $x>1$, $\operatorname{sgn} (x-x_n)=\operatorname{sgn} R_n(x)$
For all $x>1$, $R_n(x)$ converges pointwise to $Q(x)=x^2-x-1$
But $\operatorname{sgn} Q(x)=\operatorname{sgn} (x-\phi)$ where $\phi=(1+\sqrt 5)/2$, so that whenever $x\ne\phi$, $\operatorname{sgn} Q(x)\ne 0$ and we have: $\lim_{n\to\infty} \operatorname{sgn} (x-x_n)=\lim_{n\to\infty} \operatorname{sgn} R_n(x)=\operatorname{sgn} Q(x)=\operatorname{sgn} (x-\phi)$ and therefore $x_n$ converges to $\phi$.