A sequence of roots of polynomials depending on an integer parameter For $n\in \mathbb N-\{0\}$, let 
$$Q_n=(2n-1)X^n+(2n-3)X^{n-1}+(2n-5)X^{n-2}+\cdots+3X^2+X$$
I want to show that there is a unique $x_n\geq 0$ such that $Q_n(x_n)=1$ and then show that the sequence $x_n$ is convergent.
Since $Q_n(0)=0$ and $\lim_{x\to \infty}Q_n=\infty$ then it suffices to show that $Q_n$ is increasing but how to do that and how to calculate the limit of the sequence $x_n$ ? thank you for your help !! 
 A: Here is an alternative proof.
Let $P_n=Q_n-1$. Then $P_n=\sum_{j=0}^n (2j-1)X^j$. $P'_n(x)=\sum_{j=1}^n j(2j-1)x^{j-1}$
is plainly nonnegative when $x\geq 0$, so $P_n$ is increasing on $[0,+\infty)$.
Since $P_n(0)=-1<0$ and ${\lim}_{+\infty}P_n=+\infty$, we see
that there is a unique $x_n\geq 0$ such that $P_n(x_n)=0$.
Also, we have the identity (see DanielR’s answer for more explanations
on how it is obtained)
$$
P_n=\frac{(2n-1)X^{n+2}-(2n+1)X^{n+1}+3X-1}{(1-X)^2} \tag{1}
$$
whence
$$
P_n(\frac{1}{3})=\frac{9}{4}\left(\frac{1}{3}\right)^{n+1}
(-\frac{4n+4}{3})<0 \tag{2}
$$
and for $\varepsilon >0$,
$$
P_n(\frac{1}{3}+\varepsilon)=\frac{1}{(\frac{2}{3}-\varepsilon)^2}
\Bigg((2n-1)\left(\frac{1}{3}+\varepsilon\right)^{n+2}-(2n+1))\left(\frac{1}{3}+\varepsilon\right)^{n+1}+3\varepsilon\Bigg)
 \tag{3}
$$
It follows that
$$
\lim_{n\to+\infty} P_n(\frac{1}{3}+\varepsilon)=
\frac{3\varepsilon}{(\frac{2}{3}-\varepsilon)^2}>0, \tag{4}
$$
so that $x_n\in[\frac{1}{3},\frac{1}{3}+\varepsilon]$ for large enough $n$.
A: When $x>0$, the terms in $Q'_n(x)$ are all positive and so $Q'_n(x)>0$. Hence, $Q_n$ is increasing.
Now, $Q_n(x) =(2n-1)X^n+ Q_{n-1}(x)$ and so $Q_n(x_{n-1}) =(2n-1)x_{n-1}^n+1 > 1$. Hence $x_n < x_{n-1}$. Since $0 <x_n$, the sequence converges.
A: To show that $Q_n(x)$ is increasing for $x \geqslant 0$, take the derivative and note that all of the terms are positive.
Once convergence is shown (see lhf's answer), the limit can be calculated this way:
$$\begin{align}
\sum_{k=1}^\infty(2k-1)x^k&=2\sum_{k=1}^\infty kx^k-\sum_{k=1}^\infty x^k\\
&=\{x\in(0,1)\}\\
&=\frac{2x}{(x-1)^2}+\frac1{x-1} \equiv p(x)\\
\end{align} $$
Now, solving $p(x)-1=0$ gives you $x=\frac13$.
