THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Let $F:[0,\infty )\to \mathbb{R}$ be the antiderivative of $f$ vanishing at $0$. Since $f$ takes on positive values, $F$ is strictly increasing, hence injective, and $\alpha =\sup \text{im }F>0$, the supremum being considered on the extended line.

a) Fix a positive integer $n_0$ such that $n_0\alpha \ge 1$. If $n$ is an integer greater than $n_0$, then $F(x_n)=1/n$ for some $x_n>0$, by the intermediate value theorem. Uniqueness of $x_n$ follows from injectivity of $F$.

b) Since $F(x_{n+1})=1/(n+1)<1/n=F(x_n)$, and $F$ is strictly increasing, $(x_n{)}_{n>n_0}$ is a strictly decreasing sequence of positive real numbers, so it is convergent. By continuity, $F({\lim }_{n\to \infty }{x}_n)={\lim }_{n\to \infty }F(x_n)=0=F(0)$, so ${\lim }_{n\to \infty }{x}_n=0$, by injectivity of $F$. For each integer $n>n_0$, refer to the first mean value theorem to write $1/n=F(x_n)=x_{n}f(t_n)$, i.e., $n{x}_n=1/f(t_n)$, for some positive $t_n<x_n$. Since $(x_n{)}_{n>n_0}$ converges to $0$, so does $(t_n{)}_{n>n_0}$. Consequently, $(n{x}_n{)}_{n>n_0}$ converges to $1/f(0)$, by continuity of $f$.