The problem at hand is to show if $x_n$ is decreasing and $\sum x_n$ converges, then $nx_n \to 0.$

My approach was to assume the opposite, i.e. there exist infinitely many $n_j$ with $n_jx_{n_j} > \epsilon,$ and then estimate $$\sum x_n \geq \epsilon (1 + \frac{n_2-n_1}{n_2} + \frac{n_3-n_2}{n_3} + \ldots)$$ which easily follows from decreasingness. Now we see that the problem and the statement in the title are equivalent, else we could take an $a_n$ and construct a counterexample. I sadly can't see how to prove either. Thank you for your help!


1 Answer 1


Let $a_n$ be increasing and non-negative. Set $\epsilon_n = \frac{a_n - a_{n-1}}{a_n}$. Then, we have : $$\ln a_n - \ln a_{n-1} = -\ln(1-\epsilon_n) \leqslant \epsilon_n$$

Therefore, if $\sum_n \epsilon_n$ converges, we see that $a_n$ is bounded.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .