Question on the existence of a boundary between divergence and convergence It is said that $\sum1/n^{1+\epsilon}$ will converge and $\sum1/n$ will not. There are various proofs showing that such a boundary between divergence and convergence does not, in principle, exist. This is of course evident in the above, where $\epsilon$ can be made arbitrarily small, e.g. $0.1, 0.01, 0.001$, etc. Resultantly, this boundary can't really be cast in specific terms. However, I am wondering if it is permissible to nonetheless say that $1/n$ is the fastest decreasing divergent series. That is to say that anything decreasing faster, even by an infinitesimally small value, will converge. Can we refer to this 'boundary' not as a boundary per se, but a minimum condition, the minimum being the series $\sum S$ converges only if it decreases with greater speed than $\sum 1/n$?  

The question is interesting. Actually, things are not so simple. There are many (infinite) different series "between"
$$
\sum \frac{1}{n}=\infty  \quad\textrm{and}\quad  \sum \frac{1}{n^{1+\varepsilon}}<\infty
$$
(for any $\varepsilon>0$),
some convergent and some other divergent. For example:
$$
\sum \frac{1}{n\cdot \ln(n)}=\infty,  \quad 
\sum \frac{1}{n\cdot (\ln\,n)^{1+\varepsilon}}<\infty
$$
and
$$
\sum \frac{1}{n\cdot \ln(n)\cdot \ln(\ln (n))}=\infty, \quad 
\sum \frac{1}{n\cdot \ln(n)\cdot (\ln (\ln (n)))^{1+\varepsilon}}<\infty
$$
and so on.
The ordering of the series is 
$$
\frac{1}{n}>\frac{1}{n\cdot \ln(n)}>\frac{1}{n\cdot \ln(n)\cdot \ln(\ln (n))}>\ldots \\
\ldots > \frac{1}{n\cdot \ln(n)\cdot (\ln (\ln (n)))^{1+\varepsilon}}>\frac{1}{n\cdot (\ln\,n)^{1+\varepsilon}}>\frac{1}{n^{1+\varepsilon}}
$$
(For each $\varepsilon>0$, $1/n^\varepsilon<1/(\ln n)^k$ for $n$ sufficiently large.)
