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gpt pro 5.2 attempt on Math-644, version 3

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Problem 15 $v3$: Erdős alternating series $\sum_{n\ge 1} (-1)^n\, \frac{n}{p_n}$ (Round 3)
1) ROUND-3 OBJECTIVE
2) Round-2 FOUNDATION USED
3) NEW INSIGHT / TOOL (ROUND 3)
4) ATTACK PLAN (ROUND 3)
5) WORK (ROUND 3)
5.1 A family of sparse sets with controlled counting function and divergent $\sum 1/(m\log m)$
5.2 Equidistributed sequences with small discrepancy but divergent weighted sum
5.3 Consequence for discrepancy-only strategies
5.4 Back to Erdős' series
6) ADVERSARIAL VERIFICATION
7) FINAL (EXACTLY ONE)
8) COMPLETION ESTIMATE (MANDATORY)
9) REFERENCES
References
10) OUTPUT FORMAT

Problem 15 $v3$: Erdős alternating series $\sum_{n\ge 1} (-1)^n\, \frac{n}{p_n}$ (Round 3)

1) ROUND-3 OBJECTIVE

Path (C): obstruction/correction sharpened. Building on Round 2's Said--Tao reduction, the remaining obstruction is quantitative cancellation in the parity process $(-1)^{\pi(m)}$. Round 2 produced (i) an equivalence between the Erdős series and the parity series $\sum (-1)^{\pi(m)}/(m\log m)$, and (ii) a sufficient discrepancy bound $A(x)=O\left(x/(\log\log x)^{1+\varepsilon}\right)$ for convergence.

In Round 3, the most promising advance is to show this discrepancy-only approach is essentially sharp:

the power $1+\varepsilon$ of $\log\log$ in the sufficient condition cannot be reduced to $\le 1$ if one only controls discrepancy $A(x)$;
more precisely, for every $\alpha\le 1$ there are equidistributed sign sequences with $A(x)=O\left(x/(\log\log x)^\alpha\right)$ for which the weighted series $\sum \varepsilon(m)/(m\log m)$ diverges.

Thus any strategy that aims to settle Erdős by proving a bound of the form $A(x)\ll x/(\log\log x)^\alpha$ must reach an exponent $\alpha>1$ (or exploit additional structure beyond discrepancy).

2) Round-2 FOUNDATION USED

We rely on the following Round 2 results:

Said--Tao equivalence (Round 2 Theorem 5.7): \[ S(x):=\sum_{n\le x} (-1)^n\frac{n}{p_n} = \frac12\,T(x\log x)+C+o(1),\qquad T(y):=\sum_{2\le m\le y}\frac{(-1)^{\pi(m)}}{m\log m}, \] hence convergence of $\sum (-1)^n n/p_n$ is equivalent to convergence of $\sum (-1)^{\pi(m)}/(m\log m)$.
Discrepancy criterion (Round 2 Proposition 5.8): if $A(x):=\sum_{2\le m\le x}(-1)^{\pi(m)}$ satisfies \[ A(x)=O\left(\frac{x}{(\log\log x)^{1+\varepsilon}}\right) \quad\text{for some }\varepsilon>0, \] then $T(y)$ converges, hence the Erdős series converges.
Barrier example (Round 2 Proposition 5.9): there exists an equidistributed sign sequence $\varepsilon(m)\in\{\pm 1\}$ for which $\sum \varepsilon(m)/(m\log m)$ diverges.

3) NEW INSIGHT / TOOL (ROUND 3)

New tool: sharpness of the $\log\log$ exponent for discrepancy-only arguments.

Round 2 showed that a bound $A(x)=O\left(x/(\log\log x)^{1+\varepsilon}\right)$ suffices for convergence of $\sum \varepsilon(m)/(m\log m)$. Round 3 shows the exponent $1+\varepsilon$ is essentially best possible within the class of power bounds in $\log\log$:

for every $0<\alpha\le 1$, one can still have equidistribution and even the quantitative bound $A(x)=O\left(x/(\log\log x)^\alpha\right)$, while forcing divergence.

This rules out an entire class of proof attempts (those that target $A(x)=O(x/(\log\log x)^\alpha)$ with $\alpha\le 1$ and nothing deeper).

4) ATTACK PLAN (ROUND 3)

Round 2 gap to close: Proposition 5.8 gave a sufficient quantitative condition on $A(x)$, but Round 2 did not clarify whether this is in any sense minimal or sharp.

What must be proved now:

Construct, for each $\alpha\le 1$, an equidistributed $\{\pm1\}$-sequence $\varepsilon(m)$ with discrepancy $A_\varepsilon(x):=\sum_{3\le m\le x}\varepsilon(m)=O\left(x/(\log\log x)^\alpha\right)$ for which $\sum_{m\ge 3} \varepsilon(m)/(m\log m)$ diverges.

Once done, we can conclude that (within discrepancy-only strategies) one needs a power strictly larger than $1$ of $\log\log$.

5) WORK (ROUND 3)

In this section we work with an arbitrary sign sequence $\varepsilon(m)\in\{\pm1\}$. Define its discrepancy \[ A_\varepsilon(x):=\sum_{3\le m\le x}\varepsilon(m). \]

5.1 A family of sparse sets with controlled counting function and divergent $\sum 1/(m\log m)$

Lemma (Sparse divergent set with power $\log\log$ control): Fix $\alpha\in(0,1]$ and define, for $k\ge 3$, \[ n_k := 2\lfloor k\,(\log\log k)^\alpha\rfloor+1, \qquad Y:=\{n_k: k\ge 3\} \subseteq 2\mathbb{N}+1. \] Then:

(Counting function) $\#(Y\cap[1,N]) = O\left(N/(\log\log N)^\alpha\right)$.
(Divergent weight) $\sum_{m\in Y} \frac{1}{m\log m}=\infty$.

Proof: Step 1: monotonicity. For $k$ large, $k(\log\log k)^\alpha$ is strictly increasing, hence $n_k$ is strictly increasing for all $k$ beyond some absolute threshold; discarding finitely many initial terms does not affect either claim.

Step 2: counting function. Let $N$ be large and let $K:=\#(Y\cap[1,N])$, so $n_K\le N< n_{K+1}$. We split into cases.

Case 1: $K\le N^{1/2}$. Then trivially $K\le N^{1/2} \le N/(\log\log N)^\alpha$ for $N$ sufficiently large (since $(\log\log N)^\alpha=o(N^{1/2})$). This gives the desired bound.

Case 2: $K> N^{1/2}$. Then $\log\log K \ge \log\log(N^{1/2}) = \log(\log N - \log 2)$. For $N$ large this implies $\log\log K \ge \tfrac12\log\log N$. Using $n_K \ge 2\lfloor K(\log\log K)^\alpha\rfloor \ge K(\log\log K)^\alpha$ for large $K$, we obtain \[ N \ge n_K \ge K\,(\log\log K)^\alpha \ge K\,\left(\tfrac12\log\log N\right)^\alpha. \] Thus \[ K \le 2^\alpha\,\frac{N}{(\log\log N)^\alpha}. \] Combining both cases yields $\#(Y\cap[1,N])=O\left(N/(\log\log N)^\alpha\right)$.

Step 3: divergence of $\sum_{m\in Y} 1/(m\log m)$. For large $k$ we have \[ n_k \le 2k(\log\log k)^\alpha + 1 \le 3k(\log\log k)^\alpha, \qquad \log n_k \le 2\log k, \] so \[ \frac{1}{n_k\log n_k} \ge \frac{c}{k\,\log k\,(\log\log k)^\alpha} \quad\text{for some }c>0\text{ and all large }k. \] Therefore \[ \sum_{m\in Y}\frac{1}{m\log m} \ge c\sum_{k\ge k_0}\frac{1}{k\,\log k\,(\log\log k)^\alpha}. \] The last series diverges for $\alpha\le 1$ by the integral test: set $u:=\log\log k$, so $du = \frac{dk}{k\log k}$, and \[ \int^{\infty}\frac{dk}{k\log k\,(\log\log k)^\alpha} = \int^{\infty}\frac{du}{u^\alpha} =\infty \qquad (\alpha\le 1). \] This proves the divergence. ∎

5.2 Equidistributed sequences with small discrepancy but divergent weighted sum

Proposition (Sharpness of the $\log\log$-exponent barrier): Fix $\alpha\in(0,1]$. There exists an equidistributed sign sequence $\varepsilon(m)\in\{\pm1\}$ such that

$A_\varepsilon(N)=O\left(N/(\log\log N)^\alpha\right)$, and
the series $\sum_{m\ge 3} \varepsilon(m)/(m\log m)$ diverges.

Proof: Let $Y\subseteq 2\mathbb{N}+1$ be the set from Lemma lem:sparse-set. Define \[ \varepsilon(m) := \begin{cases} +1, & m\in Y,\\ (-1)^m, & m\notin Y. \end{cases} \] Step 1: discrepancy bound and equidistribution. The baseline sequence $(-1)^m$ has discrepancy bounded by $1$. Each modification at $m\in Y$ flips the sign from $(-1)^m=-1$ (since $m$ is odd) to $+1$, changing the running discrepancy by $+2$. Thus \[ |A_\varepsilon(N)| \le 1 + 2\#(Y\cap[1,N]) = O\left(\frac{N}{(\log\log N)^\alpha}\right) \] by Lemma lem:sparse-set$1$. In particular $A_\varepsilon(N)=o(N)$, so $\varepsilon$ is equidistributed.

Step 2: divergence of the weighted series. Decompose \[ \sum_{m\ge 3}\frac{\varepsilon(m)}{m\log m} = \sum_{m\ge 3}\frac{(-1)^m}{m\log m} + \sum_{m\in Y}\frac{\varepsilon(m)-(-1)^m}{m\log m}. \] The first series converges by the alternating series test (since $1/(m\log m)$ decreases to $0$). For $m\in Y\subseteq 2\mathbb{N}+1$, we have $(-1)^m=-1$ and $\varepsilon(m)=+1$, hence $\varepsilon(m)-(-1)^m=2$. Therefore the second series equals \[ 2\sum_{m\in Y}\frac{1}{m\log m}, \] which diverges by Lemma lem:sparse-set$2$. Thus the full weighted series diverges. ∎

5.3 Consequence for discrepancy-only strategies

Corollary (Critical exponent for power $\log\log$ discrepancy): Consider any sign sequence $\varepsilon(m)\in\{\pm1\}$.

If $A_\varepsilon(x)=O\left(x/(\log\log x)^{1+\delta}\right)$ for some $\delta>0$, then $\sum_{m\ge 3} \varepsilon(m)/(m\log m)$ converges.
For every $\alpha\in(0,1]$ there exists an equidistributed $\varepsilon$ with $A_\varepsilon(x)=O\left(x/(\log\log x)^\alpha\right)$ for which the same series diverges.

Hence, within the family of criteria of the form $A_\varepsilon(x)\ll x/(\log\log x)^\beta$, the threshold $\beta=1$ is sharp.

Proof: Part $1$ is exactly Round 2 Proposition 5.8 (applied to a general sign sequence; its proof uses only partial summation and does not exploit prime structure). Part $2$ is Proposition prop:sharpness. ∎

5.4 Back to Erdős' series

By Round 2 Theorem 5.7, the Erdős series converges if and only if the parity series \[ \sum_{m\ge 2}\frac{(-1)^{\pi(m)}}{m\log m} \] converges. Corollary cor:critical shows that a proof based solely on a bound \[ \sum_{m\le x}(-1)^{\pi(m)} \ll \frac{x}{(\log\log x)^\beta} \] would have to reach some $\beta>1$; any claim with $\beta\le 1$ is provably insufficient in general (even under equidistribution).

6) ADVERSARIAL VERIFICATION

Quantifiers / domains. We start at $k\ge 3$ so $\log\log k$ is defined and positive; finitely many initial terms do not affect divergence or asymptotic discrepancy.
Counting bound in Lemma lem:sparse-set$1$. The split into $K\le N^{1/2}$ and $K> N^{1/2}$ ensures a uniform lower bound $\log\log K \ge \tfrac12\log\log N$ in the latter case; constants can be adjusted without changing the $O(\cdot)$ conclusion.
Divergence in Lemma lem:sparse-set$2$. The comparison $1/(n_k\log n_k)\gg 1/(k\log k\,(\log\log k)^\alpha)$ uses only crude bounds $n_k\ll k(\log\log k)^\alpha$ and $\log n_k\ll \log k$; both are valid for all large $k$.
Discrepancy calculation. Each modified odd $m\in Y$ flips the sign from $-1$ to $+1$, changing the running sum by $+2$; therefore $|A_\varepsilon(N)|\le 1+2\#(Y\cap[1,N])$ is sharp and needs no hidden assumptions.
No interaction with prime structure claimed. Proposition prop:sharpness concerns abstract sign sequences; it is used only to certify a barrier for discrepancy-only approaches, not to assert anything about $(-1)^{\pi(m)}$.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 3 strengthens the Round 2 obstruction by showing the Round 2 discrepancy criterion is essentially sharp (in the natural power scale of $\log\log$):

$A(x)=O\left(x/(\log\log x)^{1+\varepsilon}\right)$ is sufficient for convergence (Round 2),
but for every $\alpha\le 1$ there exist equidistributed sequences with $A(x)=O\left(x/(\log\log x)^\alpha\right)$ for which $\sum \varepsilon(m)/(m\log m)$ diverges (Round 3).

Therefore any attempt to settle Erdős via a discrepancy-only bound of the form $\sum_{m\le x}(-1)^{\pi(m)} \ll x/(\log\log x)^\beta$ must reach $\beta>1$ or exploit additional structure specific to the prime-parity sequence.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 75%

9) REFERENCES

References

[TaoArxiv2023] T. Tao, The convergence of an alternating series of Erdős, assuming the Hardy--Littlewood prime tuples conjecture, arXiv:2308.07205 $2023$.
[TaoBlog2023] T. Tao, The convergence of an alternating series of Erdős, assuming the Hardy--Littlewood prime tuples conjecture, blog post (Aug 2023).
[MO313999] MathOverflow question 313999, A problem of Erdős on convergence of $\sum (-1)^n n/p_n$ and equidistribution of parity of $\pi(m)$ (Oct 2018).

10) OUTPUT FORMAT

No input x.tex file was available in this session; this writeup continues the Round 2 investigation as provided.
This file is intended to be saved as 15_v3_solutions.tex (or 15_v3.tex if your pipeline expects that name).
The content begins with \section and contains no preamble, as requested.

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@misc{wesolve:ERDOS-15-A6-r1,
  title = {{gpt pro 5.2 attempt on ERDOS-15, version 3}},
  author = {{we-solve research archive}},
  year = {2026},
  month = {1},
  howpublished = {\url{https://we-solve.github.io/contributions/erdos-problem-15-1906c690/attempt-6/r1/}},
  note = {Revision r1 of ERDOS-15-A6. Unverified research record. LLMs and submitter roles are disclosed in the record.}
}

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