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I am struggling to find a closed form of the following series:

$$ \Re\Big{(}\sum_{k=1}^\infty\frac{i^{\sigma_0(k)}}k \Big{)}$$

Where $i=\sqrt{-1}$ and $\sigma_0(k)$ is the number of divisors of $k$, relevant OEIS. I understand this is essentially a problem of 'when is $\sigma_0(k)$ equal to $2,4,$ a multiple of $2$ divisible by $4$, or a multiple of $2$ indivisible by $4$?' but I am not too knowledgable on how to pick apart $\sigma_0(k)$, which makes approaching this problem very difficult.

gen-ℤ ready to perish
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Descartes Before the Horse
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  • Who said there is a closed-form ? $i^{\sigma_0(k)}$ is not periodic, it is not defined by some congruences. – reuns Dec 06 '19 at 00:02

1 Answers1

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$$\sum_{k=1}^\infty\frac{\Re(i^{\sigma_0(k)})}k =+\infty$$ This is because $\sigma_0(k)=\prod_{p^r\| k} (r+1)$ so that $i^{\sigma_0(k)}$ is real iff some $p^{2m+1}\| k$ and it suffices that another $q^{2l+1}\| k$ to get that $i^{\sigma_0(k)}=1$ whereas $i^{\sigma_0(k)}=-1$ iff $k=p^{2m+1} n^2$ which is much rarer.

More explicit : any integer is of the form $mn^2$ with $|\mu(m)|=1$ in a unique way and $$i^{\sigma_0(mn^2)}=\cases{\not\in \Bbb{R} \iff m=1\\ =-1 \iff m \text{ is prime and } v_p(n) \text{ is even} \\ =1 \text{ otherwise}}$$ Whence for $s > 1$ $$\sum_{k=1}^\infty\frac{\Re(i^{\sigma_0(k)})}{k^s} = \sum_{m\ge 2,n\ge 1}\frac{|\mu(m)|}{(mn^2)^s}-2 \sum_{p\text{ prime},r\ge 1,n\ge 1,p\ \nmid\ n}\frac{1}{(p^{4r+1}n^2)^s}$$ $$=\zeta(2s)\left(\frac{\zeta(s)}{\zeta(2s)}-1\right)-2\sum_{p,r\ge 1}(1-p^{-2s})\zeta(2s)p^{-(4r+1)s}$$ $$ = \zeta(s)-\zeta(2s)-2\zeta(2s)\sum_{r\ge 1} (-1)^{r}P(s(2r+1))$$

reuns
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