What can be said about $a_n$?

Question

Suppose that we have some real sequence $a_n$ such that $a_n \geq 0$ and such that $\sum_{{n=1}}^{+ \infty} \dfrac {a_n}{n^{\frac {1}{k}}}$ converges for every $k \in \mathbb N$.

Do we have then $a_n=0$ for all except maybe finitely many $n \in \mathbb N$?

Answer 1

We always have $n^{\frac1k}\ge 1$ so if $\sum_{n=1}^\infty a_n$ converges, then so does $$\sum_{{n=1}}^{+ \infty} \dfrac {a_n}{n^{\frac {1}{k}}}.$$

So for example $a_n=q^n$ for $0<q<1$ will make the series converge without any terms being $0$.

Answer 2

$a_n={1\over n}$ fits your description.

Upd. I implicitly assumed that you meant $\sum a_n$ to be divergent, otherwise the question becomes too trivial. As Kusma noted, any convergent series would do.