# Asymptotic estimate for the sum $\sum_{n\leq X}\mu(n)\tau(n)$?

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.

• 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 $\Re(s) > \sigma-\epsilon$ – reuns Oct 8 '18 at 23:13