Vijetnam 2007

For every $n$ we define $g_n(x)=f_n(x)-a$. Then $g_n(x)$ is a continuous and increasing function on $[0;+\infty )$. We have $g_n(0)=1-a<0$; $g_n(1)=a^{10}+n+1-a>0$ so $g_n(x)=0$ has the only root $x_n$ in $(0;+\infty )$.

To prove the existence of the limit ${\lim }_{n\to \infty }{x}_n$, we prove that the sequence $(x_n)$, $n=1,2,\dots$, is increasing and confined.

$$\begin{array}{rl}\text{We have }& g_n\left(1-\frac{1}{a}\right)=a^{10}{\left(1-\frac{1}{a}\right)}^{n+10}+\frac{1-{\left(1-\frac{1}{a}\right)}^{n+1}}{1}-a\\ & =a{\left(1-\frac{1}{a}\right)}^{n+1}\left(a^9{\left(1-\frac{1}{a}\right)}^9-1\right)=a{\left(1-\frac{1}{a}\right)}^{n+1}\left((a-1{)}^9-1\right)>0.\end{array}$$

$$\text{Thus }{x}_n<1-\frac{1}{a}\forall n.$$

On the other hand $g_n(x_n)=a^{10}{x}_n^{n+10}+x_n^n+\cdots +1-a=0$, therefore

$$\begin{array}{rl}{x}_n{g}_n(x_n)& =a^{10}{x}_n^{n+11}+x_n^{n+1}+\cdots +x_n-a{x}_n=0\\ \Rightarrow g_{n+1}(x_n)& =x_n{g}_n(x_n)+1+a{x}_n-a=a{x}_n+1-a<0\text{ for }{x}_n<1-\frac{1}{a}.\end{array}$$

Since the function $g_{n+1}$ is increasing and $0=g_{n+1}(x_{n+1})>g_{n+1}(x_n)$ then we have $x_n<x_{n+1}$. Thus the sequence $(x_n)$, $n=1,2,\dots$, is increasing and confined, and therefore there exists ${\lim }_{n\to \infty }{x}_n$.