TITLE: Is there a slowest rate of divergence of a series?
QUESTION [66 upvotes]: $$f(n)=\sum_{i=1}^n\frac{1}{i}$$
diverges slower than
$$g(n)=\sum_{i=1}^n\frac{1}{\sqrt{i}}$$
, by which I mean $\lim_{n\rightarrow \infty}(g(n)-f(n))=\infty$. Similarly, $\ln(n)$ diverges as fast as $f(n)$, as $\lim_{n \rightarrow \infty}(f(n)-\ln(n))=\gamma$, so they 'diverge at the same speed'. 
I think there are an infinite number of 'speeds of divergence' (for example, $\sum_{i=1}^n\frac{1}{i^k}$ diverge at different rates for different $k<1$). However, is there a slowest speed of divergence?
That is, does there exist a divergent series, $s(n)$, such that for any other divergent series $S(n)$, the limit $\lim_{n \rightarrow \infty}(S(n)-s(n))=\infty$ or $=k$? If so, are there an infinite number of these slowest series?

REPLY [2 votes]: Suppose there is a slowest-diverging series. Call it $(a_n)$.
Define $A(n)=\sum a_n$. Thus, $A(n)$ is a slowly-diverging function.
Define $B(n)=\ln A(n)$. Thus, $B(n)$ diverges even slower than $A(n)$.
Define $b_n=B(n)-B(n-1)$. Thus, $(b_n)$ is an even slower diverging series. Contradiction.