, 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.

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

$$\text{Thus } x_n < 1 - \frac{1}{a} \quad \forall n.$$

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

$$\begin{aligned} x_n g_n(x_n) &= a^{10} x_n^{n+11} + x_n^{n+1} + \dots + x_n - a x_n = 0 \\ \Rightarrow g_{n+1}(x_n) &= x_n g_n(x_n) + 1 + a x_n - a = a x_n + 1 - a < 0 \text{ for } x_n < 1 - \frac{1}{a}. \end{aligned}$$

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 $\lim_{n \to \infty} x_n$.