A limit involving polynomials 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:
$$\lim_{n\to\infty} x_{n}$$
 A: $$P_n(x)=x^{n+1}-(x^{n-1}+x^{n-2}+...x+1)=x^{n+1}-\sum_{k=0}^{n-1}x^{k}$$
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?
A: 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 $\phi$.
A: 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. 

Figure 1. Contour plot of $|P_{15}(x+i y)|$.
