Mathematical Reflections 2012, Issue 6 - Problem U249

Problem:
Let $(a_n)_{n \geq 1}$ be a decreasing sequence of positive numbers. Let $$s_n=a_1+a_2+\ldots+a_n,$$ and $$b_n=\dfrac{1}{a_{n+1}}-\dfrac{1}{a_n},$$ for all $n \geq 1$. Prove that if $(s_n)_{n \geq 1}$ is convergent, then $(b_n)_{n \geq 1}$ is unbounded.

Proposed by Bogdan Enescu.

Solution:
Since $(s_n)_{n \geq 1}$ is convergent, then $\lim_{n \to \infty} a_n=0$. Suppose that $(b_n)_{n \geq 1}$ is bounded. Then there exists $M \in \mathbb{R}^+$ such that $b_n \leq M$ for all $n \geq 1$. By the Problem 2.5.12 at page 97 of Radulescu, Andreescu - Problems In Real Analysis, the sequence $(a_n)_{n \geq 1}$ converges to zero if and only if the series $$\sum_{n=1}^\infty \left(1-a_{n+1}/a_n\right)=\sum_{n=1}^\infty a_{n+1}b_n$$ diverges. But
$$\sum_{n=1}^\infty a_{n+1}b_n \leq \sum_{n=1}^\infty a_nb_n \leq M\sum_{n=1}^\infty a_n < \infty,$$ contradiction.