IMC 2000 · Day 1 · P6

From the conditions it is obvious that $F$ is increasing and $\lim\limits_{n \to \infty} b_n = \infty$.

By Lagrange's theorem and the recursion in (1), for all $k \ge 0$ integers there exists a real number $\xi \in (a_k, a_{k+1})$ such that $$\tag{2} F(a_{k+1}) - F(a_k) = f(\xi)(a_{k+1} - a_k) = \frac{f(\xi)}{f(a_k)}.$$ By the monotonity, $f(a_k) \le f(\xi) \le f(a_{k+1})$, thus $$\tag{3} 1 \le F(a_{k+1}) - F(a_k) \le \frac{f(a_{k+1})}{f(a_k)} = 1 + \frac{f(a_{k+1}) - f(a_k)}{f(a_k)}.$$ Summing (3) for $k = 0, \dots, n-1$ and substituting $F(b_n) = n$, we have $$\tag{4} F(b_n) < n + F(a_0) \le F(a_n) \le F(b_n) + F(a_0) + \sum_{k=0}^{n-1} \frac{f(a_{k+1}) - f(a_k)}{f(a_k)}.$$ From the first two inequalities we already have $a_n > b_n$ and $\lim\limits_{n \to \infty} a_n = \infty$.

Let $\varepsilon$ be an arbitrary positive number. Choose an integer $K_\varepsilon$ such that $f(a_{K_\varepsilon}) > \dfrac{2}{\varepsilon}$. If $n$ is sufficiently large, then $$\begin{align*} F(a_0) &+ \sum_{k=0}^{n-1} \frac{f(a_{k+1}) - f(a_k)}{f(a_k)} = \\ &= F(a_0) + \left( \sum_{k=0}^{K_\varepsilon - 1} \frac{f(a_{k+1}) - f(a_k)}{f(a_k)} \right) + \sum_{k=K_\varepsilon}^{n-1} \frac{f(a_{k+1}) - f(a_k)}{f(a_k)} < \tag{5} \\ &< O_\varepsilon(1) + \frac{1}{f(a_{K_\varepsilon})} \sum_{k=K_\varepsilon}^{n-1} \bigl( f(a_{k+1}) - f(a_k) \bigr) < \\ &< O_\varepsilon(1) + \frac{\varepsilon}{2} \bigl( f(a_n) - f(a_{K_\varepsilon}) \bigr) < \varepsilon f(a_n). \end{align*}$$ Inequalities (4) and (5) together say that for any positive $\varepsilon$, if $n$ is sufficiently large, $$F(a_n) - F(b_n) < \varepsilon f(a_n).$$ Again, by Lagrange's theorem, there is a real number $\zeta \in (b_n, a_n)$ such that $$\tag{6} F(a_n) - F(b_n) = f(\zeta)(a_n - b_n) > f(b_n)(a_n - b_n),$$ thus $$\tag{7} f(b_n)(a_n - b_n) < \varepsilon f(a_n).$$ Let $B$ be an upper bound for $f'$. Apply $f(a_n) < f(b_n) + B(a_n - b_n)$ in (7): $$f(b_n)(a_n - b_n) < \varepsilon \bigl( f(b_n) + B(a_n - b_n) \bigr),$$ $$\tag{8} \bigl( f(b_n) - \varepsilon B \bigr) (a_n - b_n) < \varepsilon f(b_n).$$ Due to $\lim\limits_{n \to \infty} f(b_n) = \infty$, the first factor is positive, and we have $$\tag{9} a_n - b_n < \varepsilon \frac{f(b_n)}{f(b_n) - \varepsilon B} < 2 \varepsilon$$ for sufficiently large $n$.

Thus, for arbitrary positive $\varepsilon$ we proved that $0 < a_n - b_n < 2\varepsilon$ if $n$ is sufficiently large.