Necessary and/or sufficient conditions for summability of a sequence

Question

It is clearly true that any $(a_n)_{n=1}^\infty$ that has $$a_n=O(n^{-1-\varepsilon}),$$ for some fixed $\varepsilon>0$, is absolutely summable: $$\sum\limits_{n=1}^\infty |a_n|<\infty.$$

My question is what is known in the converse direction? If $a_n$ is summable must it satisfy a $O$-condition as above? If not, is there a counter-example? Many thanks in advance.

Answer 1

Counter-example: $\frac1{(n+1) \ln(n+1)^2} \in ω(\frac1{n^{1+ε}})$ as $n \to \infty$, but $\sum_{n=1}^\infty \frac1{(n+1) \ln(n+1)^2}$ converges.

The easiest way to prove it is that $\sum_{n=2}^\infty \frac1{(n+1) \ln(n+1)^2} \le \int_1^\infty \frac1{(x+1)\ln(x+1)^2}\ dx = [-\frac1{\ln(x+1)}]_1^\infty = \frac1{\ln(2)}$.

We can extend this:

$\sum_{n=2}^\infty \frac1{n\ln(n)\ln(\ln(n))\ln(\ln(\ln(n)))}$ diverges.

$\sum_{n=2}^\infty \frac1{n\ln(n)\ln(\ln(n))\ln(\ln(\ln(n)))^2}$ converges.

Both are proven by using integration in the same manner, and the pattern continues as far as you wish, which shows a fine line between convergence and divergence.