What can we say about $a_n$ if $\sum a_n/n$ converges?

Question

Suppose that $\sum_{n=1}^{\infty} a_n/n$ converges, with $a_n \geq 0$ but not necessarily decreasing. What can we say about $a_n$?

We can't say $a_n \to 0$. (Consider $a_n=1$ for n square, 0 otherwise.) But we can say that for any $\epsilon>0$, there are an infinite number of $a_n \leq \epsilon$. Is there a name for this property? What else can we say about $a_n$?

Answer 1

We can say that $$\frac{1}{n}\sum_{k=1}^{n}a_k = \text{AM}(a_1,\ldots,a_n) \to 0. $$ (We don't even need the hypotesis $a_n\geq 0$). This is known as Kronecker's lemma and it is a consequence of summation by parts.

Answer 2

Perhaps the following result is of some interest.

Suppose that $\sum_n a_n/n$ converges, $a_n\ge 0$. Then at least one of the following holds:

1. $a_n\to 0$.
2. There are infinitely many $n\ge 0$ for which $a_{n+1}/(n+1)> a_n/n$.

So you cannot conclude $a_n\to 0$, but when this does not hold, you can instead conclude that $a_n/n$ is not eventually decreasing. In particular, this implies $a_n$ is not eventually decreasing, as $a_{n+1}>(n+1)/na_n\ge a_n$ infinitely often.

For a proof, see my other answer where that $b_n$ decreasing and $nb_n\not \to 0$ implies $\sum b_n$ diverges. The content of this answer is the contrapositive of the other answer applied to $b_n=a_n/n$.