Evaluating a sum with Mobius function $\sum_{d|n}\tau(d)\mu(d)$

Question

I'm not sure how to evaluate the sum $\sum_{d|n}\tau(d)\mu(d)$, since it is also not a convolution.

Answer 1

Assume $n \gt 1$.

If $n = \prod_{i=1}^{\omega(n)} p_i^{e_i}$ with $p_i$ prime (and distinct from $p_j$) and $e_i \gt 0$, then similar to the proof that $\sum_{d|n} \mu(d) = 0$ we get that

$$\sum_{d|n} \tau(d) \mu(d) = \binom{\omega(n)}{r} (-1)^r 2^r = (1-2)^{\omega(n)} = (-1)^{\omega(n)}$$