I was fiddling around with some numerical sums in Mathematica, and I noticed that the partial sums of the following series seemed to follow a nice pattern:
$$\sum_{n=1}^{10^k}\left( \frac{1}{n} \right)^{p(n)} \approx \frac{5}{3} \cdot 10^{2k-1},$$
where $p(n) = 1,1,-1,1,1,-1,1,1,-1,\ldots$ is periodic with period 3 (OEIS A131561).
Phrased another way, it seems as though the sequence $$ a_k = \left( \frac{3}{5\cdot 10^{2k-1}} \right) \left( \sum_{n=1}^{10^k}\left( \frac{1}{n} \right)^{p(n)} \right) $$ has limit $a_k \to 1$.
Is anyone familiar with results like these? Or perhaps even with results about asymptotic growth of divergent series in general? I was a bit surprised that the partial sums of this divergent series were so "nicely" behaved; I expected them to be rather chaotic.
Am I missing something? Is this kind of behavior a result of the particular way I've chosen my partial sums? Is there something more intricate going on here?
EDIT: I modified my sum in Mathematica a bit, and now I'm more convinced that there is something non-trivial going on. Based on experimental evidence, it seems as though I can drop the dependence on powers of 10 and the result still holds: $$ \alpha_{j} = \left( \frac{3}{5} \cdot \frac{10}{j^2} \right) \left( \sum_{n=1}^{j}\left( \frac{1}{n} \right)^{p(n)} \right) \to 1, $$ or $$ \sum_{n=1}^{j}\left( \frac{1}{n} \right)^{p(n)} \approx \frac{5}{30} \cdot j^2 $$