# Landau's theorem using nth roots

This question was asked earlier at MSE .

Let $$\omega$$(n) denote the number of distinct primes dividing $$n$$. The Mobius function is defined as $$\mu(n) = (-1)^{\omega(n)}$$ 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 other roots of unity? To focus on a specific case, let's use fourth roots and define $$f(n) = i^n$$ 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 $$\prod_p (1 + i/p)$$ does not converge since $$\sum\frac{1}{p^2} < \infty$$ while $$\sum \frac{1}{p} = \infty$$. ]

$$(2)$$ The partial sums $$S(n)$$ are known to satisfy $$|S(n)| \leq 1$$ for all $$n$$. Are the partial sums $$|T(n)|$$ also bounded by some constant independent of $$n$$? [Over the initial stretch $$1 \leq n \leq 20$$ , one finds that the max $$T$$ value is $$|T(19)| = 1.57 ...$$ ].

Thanks

• I suppose that there is a typo and you wanted to define $f(n)=i^{\omega(n)}$ instead of $f(n)=i^n$? Nov 8, 2018 at 9:00
• @ChristianBernert Yes, edited appropriately. The first 10,000 approx. values of $|T(n)|$ for $f(n)=i^{\omega(n)}$ are here: pastebin.com/kdKas96M. The values seem to be increasing generally. For third roots of unity (first 3000 approx. values of $|T(n)|$ are here: pastebin.com/ismMhVv8) the behavior is different, agreeing more with part (2) of your question. Nov 8, 2018 at 13:04

## 1 Answer

Theorem 1 in Section 6.1 of Tenenbaum's Introduction to Analytic and Probabilistic Number Theory states (after taking its $$N=0$$ for simplicity) that for any $$|z|<2$$, $$\sum_{n\le x} z^{\omega(n)} \sim \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac z{p-1} \bigg) \bigg( 1-\frac1p \bigg)^z \cdot x\, (\log x)^{z-1}.$$ The same method would show that $$\sum_{n\le x} \mu^2(n) z^{\omega(n)} \sim \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac zp \bigg) \bigg( 1-\frac1p \bigg)^z \cdot x\, (\log x)^{z-1}.$$ From here, partial summation gives (for $$z\ne 0$$) $$\sum_{n\le x} \mu^2(n) \frac{z^{\omega(n)}}n = \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac zp \bigg) \bigg( 1-\frac1p \bigg)^z \cdot \frac{(\log x)^z}z + c(z) + o(1)$$ for some constant $$c(z)$$. So when $$|z|<2$$ and $$\Re z<0$$, the sum converges to this $$c(z)$$; when $$|z|<2$$ and $$\Re z>0$$, the sum grows to infinity in modulus. When $$|z|<2$$ and $$\Re z=0$$, the sum oscillates asymptotically around a circle in the complex plane. So the interesting case $$z=i$$, strangely, the answer to your question (1) is no while the answer to your question (2) is yes!