lower and upper bound for $\sum_{k=1}^n \frac{(-1)^{\Omega(k)}}k$?

Question

Are there known any lower and upper bounds for $$ \sum_{k=1}^n \frac{(-1)^{\Omega(k)}}k, $$ where $\Omega(n)$ is the number of prime factors counting multiplicities of $n$?

Or at least is it known if it is always positive?

Answer 1

Let $\lambda(n) = (-1)^{\Omega(n)}$ and \[T(x) = \sum_{n \leq x} \frac{\lambda(n)}{n}.\] The conjecture that $T(x) \geq 0$ for all $x \geq 1$ is called Turán's conjecture (though Turán took objection to this labelling, as he did not conjecture that it was true). As Jeremy Rouse stated, Haselgrove disproved this in 1958 using methods of Ingham. The first counterexample occurs at $x = 72\,185\,376\,951\,205$ and was found by Borwein, Ferguson, and Mossinghoff in 2008. This latter paper also proves the fact that \begin{align*} \liminf_{x \to \infty} \sqrt{x} T(x) & < 0, \\ \limsup_{x \to \infty} \sqrt{x} T(x) & > 0, \end{align*}

It is not known if \begin{align*} \liminf_{x \to \infty} \sqrt{x} T(x) & = -\infty, \\ \limsup_{x \to \infty} \sqrt{x} T(x) & = \infty, \end{align*} though both are conjectured to be true and, by Ingham's work, would follow from the linear independence hypothesis, which states that the positive ordinates of the imaginary parts of the nontrivial zeroes of the Riemann zeta function are linearly independent over the rationals.

As to the true size of $T(x)$, it is hard to say. Unconditionally, it is not difficult to prove that $T(x) = o(1)$ (which is in fact equivalent to the prime number theorem), while the Riemann hypothesis implies that $T(x) = O_{\varepsilon}\left(x^{-1/2 + \varepsilon}\right)$ for every $\varepsilon > 0$. A probabilistic argument of Ng gives a conjectured size of maximal growth of the summatory function of the Möbius function, and the same methods would suggest that \begin{align*} 0 > \liminf_{x \to \infty} \frac{\sqrt{x} T(x)}{\left(\log \log \log x\right)^{5/4}} & > -\infty, \\ 0 < \limsup_{x \to \infty} \frac{\sqrt{x} T(x)}{\left(\log \log \log x\right)^{5/4}} & < \infty. \end{align*}