Problem in solving a question related to sequence.

The question is :

Prove that the equation $x^{n}+x^{n-1}+\dotsb +x-1 = 0$ has exactly one positive root for all $n \in N$ and if $\{b_n\}$ is the sequence of all positive roots of these equations, then prove that the sequence is convergent and it converges to $1/2$.

I proved that the first part by Descartes' rule of signs. It is hard for me to show the convergence of the sequence. It is even complicated for me to find the general term or $n$-th term of the sequence.

Solution 1:

The equality is equivalent to $$\frac{x - x^{n+1}}{1-x}=1$$ or $$2 x -1 = x^{n+1}$$

Now, notice that $0<x< a_2 < 1$, so $$0 < 2x-1< \frac{1}{a_2^{n+1}}$$ so $$0 < x-\frac{1}{2} < \frac{1}{2 a_2^{n+1}}$$

Solution 2:

I suppose that you know the unique positive root of $x^n+x^{n-1}+\cdots +x-1=0$ is between $0$ and $1$, which means $|b_n| < 1$ for all $n\ge 2$. By the definition of $b_n$, $$ b_n^n + b_n^{n-1}+ \cdots + b_n =1 $$ and so $$ b_{n+1}^{n+1}+ b_n^n + b_n^{n-1}+ \cdots + b_n > 1 = b_{n+1}^{n+1}+b_{n+1}^n +\cdots + b_{n+1}. $$ Thus \begin{align} 0 &< b_n^n - b_{n+1}^n + b_n^{n-1}-b_{n+1}^{n-1}+\cdots + b_n - b_{n+1}\\ &=(b_n-b_{n+1})(...) \end{align} so $b_n > b_{n+1}$ for all $n\in \mathbb{N}$. $(b_n)$ is bounded and monotonously decreases, so $(b_n)$ converges by monotone convergence theorem. Let $\lim_{n\to\infty} b_n = b$. $$ b_n^n + b_n^{n-1}+ \cdots + b_n = \frac{b_n ( b_n^n -1)}{b_n-1} $$ and so $$ \frac{-b}{b-1}=1. $$ Therefore $b=\frac{1}{2}$.