Does the sequence of positive roots of $x^n+x^{n-1}+\cdots+x-1=0$ converge? For a positive integer $n$, let $a_n$ denote the unique positive real root of the equation
$$x^n+x^{n-1}+\cdots+x-1=0.$$
Then what can we say about the sequence $\{a_n\}$?
We can observe that all $a_n<1$ so if $\lim a_n$ exists, let's say at $L$, then $L \le 1$. Also
\begin{align} 
&&x^n+x^{n-1}+\cdots+x-1 &= 0 \\ 
\implies&& 
x^{n+1}-1 &= 2(x-1) \\ 
\implies&& x^{n+1} &= 2x-1 \\ 
\implies&& a_n^{n+1} &= 2a_n-1. 
\end{align}
Taking limit both side we get $0=2L-1$, i.e.
$L=\frac12$ but how to show that the limit always exists?
 A: Clearly this unique positive $x$ lies in the interval $\big(\frac12,1\big)$. Also,
$$
x^n+\cdots+x-1=0 \quad\Longrightarrow\quad (x-1)(x^n+\cdots+x)-x+1=0
\quad\Longrightarrow\quad x^{n+1}-2x+1=0.
$$
Since $x$ depends on $n$, we denote it as $x_n$.
If we set $x_n=\frac12+u_n$, then $u_n\in \big(0,\frac12\big)$, and $u_n$ satisfies
$$
f(u_n)=\Big(\frac12+u_n\Big)^{n+1}-2u_n=0.
$$
Clearly, $f(0)>0$, while $f\big(\frac1{n+1}\big)<0,$ for $n$ sufficiently large, since
$$
f\Big(\frac1{n+1}\Big)=\Big(\frac12+\frac1{n+1}\Big)^{n+1}-\frac{2}{n+1}
\\=2^{-n-1}\Bigg(\Big(1+\frac{2}{n+1}\Big)^{n+1}-\frac{2^{n+2}}{n+1}\Bigg)
<2^{-n-1}\Big(e^2-\frac{2^{n+2}}{n+1}\Big)<0,
$$
and $\frac{2^{n+2}}{n+1}\to\infty$. Thus, for $n$ large enough, $f(u)$ vanishes for some $u\in \big(0,\frac1{n+1}\big)$.
Thus $0<u_n<\frac{1}{n+1}$,
and hence, for $n$ sufficiently large
$$
\frac{1}{2}<x_n<\frac{1}{2}+\frac{1}{n+1}
$$
which implies that $\lim_{n\to\infty}x_n=\frac{1}{2}$.
A: If $x_n^n + x_{n}^{n-1} + ... +x_{n} = 1$  and
$x_{n+1} ^{n+1} + ... + x_{n+1} = 1$ then we must have $x_{n+1} \leq x_n$.
Otherwise, i.e. if $x_{n+1} > x_n$ we would have
$1 = x_{n+1}^{n+1} + x_{n+1}^{n} + ... + x_{n+1} > x_{n+1}^{n} + ... + x_{n+1} >
x_{n}^{n} + ... + x_{n} = 1$.
So, the sequence of solutions is decreasing and bounded from below by zero hence it is convergent.
