Sure. In fact, I could find a series that diverges infinitely more slowly, and then I could find another that diverges infinitely more slowly than that, and so, on, ad infinitum:
<br />
\begin{array}{l}<br />
\sum _{k=n}^{\infty } 1 \\<br />
\sum _{k=n}^{\infty } \frac{1}{k} \\<br />
\sum _{k=n}^{\infty } \frac{1}{k \log (k)} \\<br />
\sum _{k=n}^{\infty } \frac{1}{k \log (k) \log (\log (k))} \\<br />
\sum _{k=n}^{\infty } \frac{1}{k \log (k) \log (\log (k)) \log (\log (\log (k)))} \\<br />
...<br />
\end{array}<br />
But those would be contrived series, made up just for the purpose of diverging slowly. The harmonic series is about as slowly diverging a series as you're likely to bump into, unless you go hunting for slowly diverging series.
I also had in mind the point Mute made: considering just series with terms of the form ip, p=-1 is the edge case.