Let $\omega(n)$ denote the number of distinct primes dividing $n$. Then the Mobius function $\mu(n)$ is defined by $\mu(n) = (-1)^{\omega(n)}$ if $n$ is squarefree and $\mu(n) = 0$ otherwise. $\,$ Next let $S(n) = \sum_{k=1}^{n} \frac{\mu(k)}{k}$ . It is known that $S(n)$ approaches zero as $n$ approaches infinity and that this is equivalent to the prime number theorem (von Mangoldt, Landau).

What happens if we replace powers of $-1$ in the Mobius function with powers of other roots of unity? $\;$ For a specific case to focus on , let's use fourth roots and define $f(n) = i^{\omega(n)}$ if $n$ is squarefree and $f(n)=0$ otherwise. Then $f(n)$ is a multiplicative function whose initial values are $(1,i,i,0,i,-1,i,0,0,-1, ... )$. Finally, let $T(n) = \sum_{k=1}^n$ $\frac{f(k)}{k}$. $\,$ Does $T(n)$ have properties analogous to those of $S(n)$?

Questions: (1) $\,$ Does $\sum_{k=1}^\infty$ $\frac{f(k)}{k}$ converge? $\,$ [The corresponding infinite product $\prod (1+ i/p)$ does not converge since $\sum\frac1{p^2}< \infty$ but $\sum \frac1p = \infty$ ].

(2) The partial sums $S(n)$ are known to satisfy $|S(n)| \le 1$ for all $n$. Are the partial sums $|T(n)|$ likewise bounded by some constant independent of $n$ ? $\,$[For the initial stretch $1\le n \le 20$, the max $T$ value was $|T(19)|=1.57\cdots $].


  • $\begingroup$ Seems like $\pi/2$... $\endgroup$ – Kenta S Jul 5 '18 at 22:14


$$|\sum_{k=1}^{\infty} \frac{f(k)}{k}|=|\prod_{p:prime} (1+\frac{i}{p})|=\prod_{p:prime}|1+\frac{i}{p}|=\sqrt{\prod_{p:prime}(1+\frac{1}{p^2})}=\sqrt{\frac{\prod_{p:prime}(1-p^{-2})^{-1}}{\prod_{p:prime}(1-p^{-4})^{-1}}}=\sqrt{\frac{\zeta(2)}{\zeta(4)}}=\frac{\sqrt{15}}{\pi}<\infty$$ Here, $\zeta(-)$ is the Riemann zeta function.

  • $\begingroup$ At the third equals sign, you forgot a square root, $\bigl\lvert 1 + \frac{i}{p}\bigr\rvert = \sqrt{1 + \frac{1}{p^2}}$. However, while the absolute value of the partial products converges (to $\frac{\sqrt{15}}{\pi}$), the product itself diverges, as noted in the question. The partial products approach a limiting circle. I don't know whether the series converges, that looks like a difficult question. $\endgroup$ – Daniel Fischer Jul 6 '18 at 9:32

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