nt.number theory – Why is it at all reasonable that $0leqsum_{Ssubseteq[Q]^*}frac{(-1)^{|S|}}{mathrm{lcm}(S)}{|S|choose (|S|+sum_{iin S}mu(i))/2}leq1$?

Let $(Q)^*$ denote the set of all squarefree natural numbers $leq Q$, and let $mu(n)$ be the Mobius function. I recently discovered the identity

$$0leqsum_{Ssubseteq(Q)^*}frac{(-1)^{|S|}}{mathrm{lcm}(S)}{|S|choose left(|S|+sum_{iin S}mu(i)right)/2}leq1tag{1}$$

And it makes absolutely no sense to me. To facilitate discussion, we let $|hat{S}|:=frac{1}{2}|S|+frac{1}{2}sum_{iin S}mu(i)$. The reason I came upon this sum is that I was studying the function $A_Q(n):=sum_{substack{d|n \ d<Q}}mu(d)$, and I noticed the the probabilities

$$Pr_{ninmathbb{N}}(A_Q(n)=j)$$

seemed to converge as $Qtoinfty$, which is not at all obvious at first glace. Since $A_{Q+1}(n)$ and $A_Q(n)$ disagree if and only if $Q$ is squarefree and $Q|n$, and in this case $A_{Q+1}(n)=A_Q(n)+mu(Q)$, we notice that

$$Pr_{ninmathbb{N}}(A_{Q+1}(n)=j)=Pr_{ninmathbb{N}}(A_Q(n)=j)-frac{1}{Q}Pr_{ninmathbb{N}}(A_Q(Qn)=j)+frac{1}{Q}Pr_{ninmathbb{N}}(A_Q(Qn)=j-mu(Q))tag{2}$$

Using a slightly more general version of this recurrence formula along with the fact that $A_1(n)=0$ uniformly gets us that

$$Pr_{ninmathbb{N}}(A_Q(n)=j)=sum_{Ssubseteq(Q-1)^*}frac{(-1)^{|S|+j}}{mathrm{lcm}(S)}{|S|choose |hat{S}|-j}tag{3}$$

As a direct consequence, we can plug in $j=0$ to get (1). Which this proof is sound and satisfactory, it makes me wonder how on earth such cancelling can occur. Generally in these types of situations one can find groups of terms that pair up with each other and cancel, but I cannot find any method that yields a result better than the trivial bound.

If anybody has any insights about where the cancellation in the sums (2) is coming from, or even better a way of showing that they converge, then that would be greatly appreciated.