Suppose $a_n\geq 0 $ and $ \sum_{n=1}^{\infty}a_n<\infty $. Does it follow that $ a_n=o(1/n) $, as $ n\to \infty $?

Question

This question came up after going through the literature on sequences and sums. Under what conditions can we obtain this $1/n$ bound ?

Thanks in advance !

Answer 1

It does not follow. Consider $$ a_n=\cases{\frac1n& if $n$ is a perfect square\\0& otherwise} $$which has finite sum, but is not $o(1/n)$.

An assumption like $a_n$ being a monotonic sequence is sufficient for $o(1/n)$, but I don't know any good (as in useful and / or interesting) sufficient and necessary conditions.

Answer 2

You could have something like $$a_n=\begin{cases}1/k^2&n=k^3\\0&\text{otherwise}\end{cases}$$ Then $\sum a_n=\frac{\pi^2}6$, but $a_n\ne o(1/n)$; in fact, $a_n=\Omega(n^{-2/3})$.

The situation is different if the sequence $a_n$ is monotonic.