Just trying to figure out what would be the asymptotic relation for the expression

$\sum_{n\leq X}\mu(n)\tau(n)$,

where $\tau$ corresponds to the number of divisors function (often named $\sigma_0$ or just $d$).

I would like to apply it to get an asymptotic formula for the following:

$\displaystyle S_q(X)=\sum_{n\leq X, (n,q)=1}\dfrac{\mu(n)\tau(n)}{n}$?

The coprimality condition adds of course an extra difficulty, but at least I'd like to have an asymptotic for $q=1$, say.

Thanks in advance!

  • $\begingroup$ The corresponding DIrichlet series is $\prod_p (1-2 p^{-s})\approx \frac{1}{\zeta(s)^2}$ whose location of the "dominating poles" depend on the Riemann hypothesis, so all you can do for $\sum_{n \le x} \mu(n) \tau(n)$ is an explicit formula in term of the zeros and bounds similar to those for $\log(x)\sum_{n \le x} \mu(n)$. There is a simple asymptotic iff there is at least one and finitely many non-trivial zeros of $\zeta(s)$ on $\Re(s) > \sigma-\epsilon$ $\endgroup$ – reuns Oct 8 '18 at 23:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.