Convergence of a generic series

Question

So I have a sequence $\epsilon(n)$ that converge to 0 when $n$ goes to infinity. What can I say about the series for $n \rightarrow \infty$ $$\sum{\frac{\epsilon(n)}{n}}?$$ Thanks for your help.

Answer 1

Nothing. The series will converge if $\epsilon(n) = 1/n$, and will diverge if $\epsilon(n) = 1 / \log(n)$.

Answer 2

If $\lim_{n\to \infty}\epsilon(n)=0$, the series could (i) converge absolutely, (ii) diverge, or (iii) converge conditionally.

If $\epsilon(n)=\frac1n$, then the series $\sum_{n=1}^\infty \frac{\epsilon(n)}{n}=\sum_{n=1}^\infty \frac{1}{n^2}$ is seen to converge absolutely by using, for example, the integral test.

If $\epsilon(n)=\frac{1}{\log(n)}$, then the series $\sum_{n=1}^\infty \frac{\epsilon(n)}{n}=\sum_{n=1}^\infty \frac{1}{n\log(n)}$ is seen to diverge, by using, again for example, the integral test.

And if $\epsilon(n)=\frac{(-1)^n}{\log(n)}$, then the series $\sum_{n=1}^\infty \frac{\epsilon(n)}{n}=\sum_{n=1}^\infty \frac{(-1)^n}{n\log(n)}$ converges conditionally using Leibniz's test.