gpt pro 5.4 attempt on Math-644, version 2

Problem 15: Erdős' alternating prime series $\sum_{n\ge 1} (-1)^n\,\frac{n}{p_n}$ (Round 4)

1) ROUND-4 OBJECTIVE

Path (C): obstruction/correction, strengthened to an exact threshold theorem.

Round 3 proved a power-scale barrier: for discrepancy bounds of the form \[ A(x):=\sum_{m\le x} \varepsilon(m)=O\!\left(\frac{x}{(\log\log x)^\alpha}\right), \] the exponent $\alpha>1$ suffices for convergence of $\sum \varepsilon(m)/(m\log m)$, while every $\alpha\le 1$ is insufficient in general.

The natural next step is to remove the special choice $x/(\log\log x)^\alpha$ and classify all monotone discrepancy gauges. That is what this round does: for any eventually nondecreasing sublinear function $F$, I prove an exact criterion for when a bound \[ A(x)=O(F(x)) \] forces convergence of \[ \sum_{m\ge 3}\frac{\varepsilon(m)}{m\log m}. \] This gives a sharp obstruction for every "discrepancy-only" strategy, not just the power family from Round 3.

2) Round-3 FOUNDATION USED

I use the following Round-3 results without re-proving them.

1. Round 3, §5.4 (using Round 2 Theorem 5.7): the Erdős series \[ \sum_{n\ge 1}(-1)^n\frac{n}{p_n} \] converges if and only if the parity series \[ \sum_{m\ge 2}\frac{(-1)^{\pi(m)}}{m\log m} \] converges.
2. Round 3, Proposition 5.9 / Corollary 5.3: the "sparse odd-set modification of the alternating sequence" gives divergent examples with controlled discrepancy in the special family $F(x)=x/(\log\log x)^\alpha$.

The new round generalizes that special-family construction to arbitrary monotone sublinear gauges $F$.

3) NEW INSIGHT / TOOL (ROUND 4)

New tool: an exact threshold theorem for discrepancy-only criteria.

Let $w_n:=1/(n\log n)$ and let \[ A_\varepsilon(x):=\sum_{3\le m\le x}\varepsilon(m), \qquad \varepsilon(m)\in\{\pm 1\}. \] I will prove:

For every eventually nondecreasing function $F:\{3,4,\dots\}\to (0,\infty)$ with $F(n)=o(n)$, \[ \sum_{n\ge 3}\frac{F(n)}{n^2\log n}<\infty \] is exactly the threshold for whether the implication \[ A_\varepsilon(x)=O(F(x)) \implies \sum_{n\ge 3}\frac{\varepsilon(n)}{n\log n}\text{ converges} \] holds for all sign sequences $\varepsilon$.

Thus Round 2's sufficient criterion and Round 3's power-family sharpness are special cases of a single exact theorem.

4) ATTACK PLAN (ROUND 4)

Exact gap left by Round 3. Round 3 only handled the one-parameter family $F(x)=x/(\log\log x)^\alpha$. The remaining gap is to determine whether there is a general discrepancy-only criterion, and if so, what its exact threshold is.

Claims to prove now.

1. Universal sufficiency: if $\sum F(n)/(n^2\log n)$ converges and $A_\varepsilon(x)=O(F(x))$, then $\sum \varepsilon(n)/(n\log n)$ converges.
2. Universal counterexample: if $\sum F(n)/(n^2\log n)$ diverges, then there exists an equidistributed sign sequence $\varepsilon$ with $A_\varepsilon(x)=O(F(x))$ for which $\sum \varepsilon(n)/(n\log n)$ diverges.

Together these produce a complete obstruction theorem for discrepancy-only approaches.

5) WORK (ROUND 4)

5.1 Setup

Fix a sign sequence $\varepsilon(n)\in\{\pm 1\}$ for $n\ge 3$, and define its discrepancy function \[ A_\varepsilon(x):=\sum_{3\le m\le x}\varepsilon(m). \] Let \[ w_n:=\frac{1}{n\log n}\qquad (n\ge 3). \] Then $w_n$ is positive, decreasing, and $w_n\to 0$.

5.2 Universal sufficiency by Abel summation

Lemma 5.1 (Abel criterion for weighted sign series).

Let $F:\{3,4,\dots\}\to (0,\infty)$ be eventually nondecreasing and satisfy $F(n)=O(n)$. If \[ \sum_{n\ge 3}\frac{F(n)}{n^2\log n}<\infty, \] and if $A_\varepsilon(x)=O(F(x))$, then the series \[ \sum_{n\ge 3}\frac{\varepsilon(n)}{n\log n} \] converges.

Proof. For integers $3\le M\le N$, Abel summation gives \[ \sum_{n=M}^{N}\varepsilon(n)w_n = A_\varepsilon(N)w_N - A_\varepsilon(M-1)w_M + \sum_{n=M}^{N-1} A_\varepsilon(n)\,(w_n-w_{n+1}). \tag{5.1} \]

We claim that \[ 0<w_n-w_{n+1}\ll \frac{1}{n^2\log n} \qquad (n\ge 3). \tag{5.2} \] Indeed, with $f(x)=1/(x\log x)$, the mean value theorem gives \[ w_n-w_{n+1} = -f'(\xi_n) \] for some $\xi_n\in(n,n+1)$, and \[ -f'(x)=\frac{\log x+1}{x^2\log^2 x} \ll \frac{1}{x^2\log x} \] for $x\ge 3$, proving $5.2$.

Now assume $|A_\varepsilon(n)|\le C F(n)$ for all large $n$. Since $F(n)=O(n)$, \[ |A_\varepsilon(N)w_N| \ll \frac{F(N)}{N\log N}\to 0, \qquad |A_\varepsilon(M-1)w_M| \ll \frac{F(M)}{M\log M}\to 0. \tag{5.3} \] Also, by $5.2$, \[ \sum_{n=M}^{\infty}|A_\varepsilon(n)|\,(w_n-w_{n+1}) \ll \sum_{n=M}^{\infty}\frac{F(n)}{n^2\log n}\to 0 \qquad (M\to\infty) \tag{5.4} \] by hypothesis.

Thus the right-hand side of $5.1$ tends to $0$ uniformly in $N\ge M$ as $M\to\infty$, so the tails of $\sum \varepsilon(n)w_n$ are Cauchy. Hence the series converges. $\square$

5.3 Dyadic condensation for the threshold series

Lemma 5.2 (Dyadic condensation).

Let $F:\{3,4,\dots\}\to (0,\infty)$ be eventually nondecreasing. Then \[ \sum_{n\ge 3}\frac{F(n)}{n^2\log n}<\infty \quad\Longleftrightarrow\quad \sum_{k\ge 2}\frac{F(2^k)}{2^k k}<\infty. \] The same equivalence holds with "$<\infty$" replaced by "$=\infty$".

Proof. For $k\ge 2$, set \[ B_k:=\sum_{n=2^k}^{2^{k+1}-1}\frac{F(n)}{n^2\log n}. \] Because $F$ is nondecreasing, for $2^k\le n<2^{k+1}$ we have \[ F(2^k)\le F(n)\le F(2^{k+1}). \] Also there are exactly $2^k$ integers in the block $[2^k,2^{k+1})$, and throughout that block \[ 2^k\le n<2^{k+1}, \qquad k\log 2\le \log n\le (k+1)\log 2. \] Hence \[ B_k \ge 2^k\cdot \frac{F(2^k)}{2^{2k+2}(k+1)\log 2} = c_1\frac{F(2^k)}{2^k k} \tag{5.5} \] for some absolute constant $c_1>0$, and similarly \[ B_k \le 2^k\cdot \frac{F(2^{k+1})}{2^{2k}k\log 2} = c_2\frac{F(2^{k+1})}{2^k k} \le c_3\frac{F(2^{k+1})}{2^{k+1}(k+1)} \tag{5.6} \] for absolute constants $c_2,c_3>0$.

Therefore $B_k$ is squeezed between constant multiples of two consecutive dyadic terms. Summing over $k$ proves that \[ \sum_{n\ge 3}\frac{F(n)}{n^2\log n} = \sum_{k\ge 2} B_k \] converges if and only if \[ \sum_{k\ge 2}\frac{F(2^k)}{2^k k} \] converges. $\square$

5.4 Universal counterexample when the threshold diverges

Proposition 5.3 (General sparse-block counterexample).

Let $F:\{3,4,\dots\}\to (0,\infty)$ be eventually nondecreasing and satisfy $F(n)=o(n)$. Assume that \[ \sum_{n\ge 3}\frac{F(n)}{n^2\log n}=\infty. \] Then there exists an equidistributed sign sequence $\varepsilon(n)\in\{\pm 1\}$ such that \[ A_\varepsilon(x)=O(F(x)) \] and yet \[ \sum_{n\ge 3}\frac{\varepsilon(n)}{n\log n} \] diverges.

Proof. By Lemma 5.2, \[ \sum_{k\ge 2}\frac{F(2^k)}{2^k k}=\infty. \tag{5.7} \] Fix a constant $c\in(0,1/10)$ and define \[ G_k:=\lfloor cF(2^k)\rfloor \qquad (k\ge 2). \] Since $F$ is eventually nondecreasing, the sequence $(G_k)$ is eventually nondecreasing. Also, because $F(n)=o(n)$, \[ G_k=o(2^k). \tag{5.8} \] After discarding finitely many initial $k$, we may assume for every $k\ge k_0$ that \[ G_k\le 2^{k-2}. \tag{5.9} \]

Define the increments \[ M_{k_0}:=G_{k_0}, \qquad M_k:=G_k-G_{k-1} \quad (k>k_0). \] Then $M_k\ge 0$ for all $k\ge k_0$.

For each $k\ge k_0$, let \[ I_k:=[2^k,2^{k+1})\cap \mathbb{N}. \] Choose exactly $M_k$ odd integers from $I_k$; this is possible by $5.9$, because $I_k$ contains $2^{k-1}$ odd integers. Let $Y$ be the union of all chosen odd integers.

Now define \[ \varepsilon(n):= \begin{cases} +1, & n\in Y,\\ (-1)^n, & n\notin Y. \end{cases} \tag{5.10} \] So we start from the alternating sequence $(-1)^n$ and flip the sign at selected odd positions.

Step 1: discrepancy bound.

Let $N\ge 2^{k_0}$ and let $K$ be such that $2^K\le N<2^{K+1}$. The baseline sequence $(-1)^n$ has discrepancy bounded by $1$, and each modified odd index changes the partial sum by exactly $+2$. Therefore \[ |A_\varepsilon(N)| \le 1 + 2\#(Y\cap[1,N]). \tag{5.11} \] But by construction, \[ \#(Y\cap[1,N])\le \sum_{k_0\le j<K}M_j + M_K = G_K. \tag{5.12} \] Since $F$ is nondecreasing and $N\ge 2^K$, \[ G_K\le cF(2^K)\le cF(N). \tag{5.13} \] Hence \[ A_\varepsilon(N)=O(F(N)). \tag{5.14} \] Because $F(N)=o(N)$, we also get \[ A_\varepsilon(N)=o(N), \tag{5.15} \] so the sequence $\varepsilon$ is equidistributed.

Step 2: divergence of the weighted series.

We decompose \[ \sum_{n\ge 3}\frac{\varepsilon(n)}{n\log n} = \sum_{n\ge 3}\frac{(-1)^n}{n\log n} + \sum_{n\in Y}\frac{\varepsilon(n)-(-1)^n}{n\log n}. \tag{5.16} \] The first series converges by the alternating series test. For every $n\in Y$ we have $n$ odd, hence $(-1)^n=-1$ and $\varepsilon(n)=+1$, so \[ \varepsilon(n)-(-1)^n = 2. \tag{5.17} \] Thus the second series equals \[ 2\sum_{n\in Y}\frac{1}{n\log n}. \tag{5.18} \] It remains to show that this diverges.

For $n\in I_k$ one has $n<2^{k+1}$ and $\log n\le (k+1)\log 2$, so every chosen $n\in Y\cap I_k$ satisfies \[ \frac{1}{n\log n} \ge \frac{1}{2^{k+1}(k+1)\log 2} \gg \frac{1}{2^k k}. \tag{5.19} \] Therefore \[ \sum_{n\in Y}\frac{1}{n\log n} \gg \sum_{k\ge k_0}\frac{M_k}{2^k k}. \tag{5.20} \] Set \[ q_k:=\frac{1}{2^k k}. \] Then summation by parts gives, for $K\ge k_0$, \[ \sum_{k=k_0}^{K} M_k q_k = G_K q_K - G_{k_0-1}q_{k_0} + \sum_{k=k_0}^{K-1} G_k(q_k-q_{k+1}). \tag{5.21} \] Now \[ q_k-q_{k+1} = \frac{1}{2^k k}-\frac{1}{2^{k+1}(k+1)} = \frac{k+2}{2^{k+1}k(k+1)} \ge \frac12\,q_k, \tag{5.22} \] and by $5.8$, \[ G_K q_K \le \frac{cF(2^K)}{2^K K}\to 0. \tag{5.23} \] So if \[ \sum_{k\ge k_0} G_k q_k = \infty, \tag{5.24} \] then $5.21$--$5.23$ imply \[ \sum_{k\ge k_0}\frac{M_k}{2^k k}=\infty. \tag{5.25} \] Finally, \[ G_k \ge cF(2^k)-1, \tag{5.26} \] and since $\sum 1/(2^k k)<\infty$, equation $5.7$ implies \[ \sum_{k\ge k_0}\frac{G_k}{2^k k}=\infty. \tag{5.27} \] Hence $5.24$ holds, so by $5.25$, $5.20$, and $5.18$, the series $\sum \varepsilon(n)/(n\log n)$ diverges. $\square$

5.5 Exact threshold theorem

Theorem 5.4 (Exact threshold for discrepancy-only methods).

Let $F:\{3,4,\dots\}\to (0,\infty)$ be eventually nondecreasing and satisfy $F(n)=o(n)$. Then the following are equivalent:

1. \[ \sum_{n\ge 3}\frac{F(n)}{n^2\log n}<\infty. \]
2. For every sign sequence $\varepsilon(n)\in\{\pm 1\}$ with $A_\varepsilon(x)=O(F(x))$, the weighted series \[ \sum_{n\ge 3}\frac{\varepsilon(n)}{n\log n} \] converges.

Moreover, if it:threshold-sum fails, then there exists an equidistributed sign sequence $\varepsilon(n)$ with $A_\varepsilon(x)=O(F(x))$ for which the weighted series diverges.

Proof. The implication it:threshold-sum$\Rightarrow$it:all-signs is Lemma 5.1. The converse obstruction is Proposition 5.3. $\square$

5.6 Return to Erdős problem 15

Define the prime-parity discrepancy \[ A_\pi(x):=\sum_{3\le m\le x}(-1)^{\pi(m)}. \] Round 3 (via Round 2 Theorem 5.7) established that the Erdős series converges if and only if \[ \sum_{m\ge 3}\frac{(-1)^{\pi(m)}}{m\log m} \] converges. Combining that fixed Round-3 foundation with Theorem 5.4 yields the following corrected formulation.

Corollary 5.5 (Minimal corrected discrepancy criterion).

Let $F$ be eventually nondecreasing with $F(n)=o(n)$.

1. If \[ A_\pi(x)=O(F(x)) \quad\text{and}\quad \sum_{n\ge 3}\frac{F(n)}{n^2\log n}<\infty, \] then the Erdős series \[ \sum_{n\ge 1}(-1)^n\frac{n}{p_n} \] converges.
2. If \[ \sum_{n\ge 3}\frac{F(n)}{n^2\log n}=\infty, \] then a proof of convergence based only on a discrepancy bound of the form $A_\pi(x)=O(F(x))$ cannot be valid in general: there exist equidistributed sign sequences with the same size bound whose weighted series diverges.

This is the exact barrier statement sought in this round.

5.7 Concrete refinement beyond Round 3

Round 3 treated only the family $F(x)=x/(\log\log x)^\alpha$. Theorem 5.4 gives many sharper examples immediately.

Corollary 5.6 (A stricter log--log--log threshold).

Fix $r\in\mathbb{R}$ and, for $x$ large, define \[ F_r(x):=\frac{x}{(\log\log x)(\log\log\log x)^r}. \] Then:

1. If $r>1$, every sign sequence $\varepsilon(n)$ with \[ A_\varepsilon(x)=O(F_r(x)) \] has convergent weighted series $\sum \varepsilon(n)/(n\log n)$.
2. If $r\le 1$, there exists an equidistributed sign sequence $\varepsilon(n)$ with \[ A_\varepsilon(x)=O(F_r(x)) \] for which $\sum \varepsilon(n)/(n\log n)$ diverges.

Proof. Apply Theorem 5.4 with $F=F_r$. The threshold series is \[ \sum_{n\ge 3}\frac{F_r(n)}{n^2\log n} \asymp \sum_{n\ge 3}\frac{1}{n\log n\log\log n\,(\log\log\log n)^r}, \] which converges if and only if $r>1$ by the integral test with the substitutions $u=\log\log n$ and $v=\log u$. $\square$

This strict refinement was not visible in Round 3, so it is genuine new progress.

6) ADVERSARIAL VERIFICATION

• Did I accidentally assume more regularity than stated? No. The threshold theorem uses only: (i) eventual monotonicity of $F$, for dyadic condensation and counting control; and (ii) $F(n)=o(n)$, for equidistribution and to guarantee enough room in each dyadic block.
• Is Lemma 5.2 valid without a doubling hypothesis? Yes. The upper bound uses the shifted endpoint value $F(2^{k+1})$, but this only shifts the dyadic series by one index and a bounded factor; no doubling condition is required.
• Could the floor in $G_k=\lfloor cF(2^k)\rfloor$ destroy divergence? No. Since $G_k\ge cF(2^k)-1$ and $\sum 1/(2^k k)<\infty$, subtracting the floor error cannot change divergence of $\sum G_k/(2^k k)$.
• Are there enough odd numbers in each block? Yes. Since $F(2^k)=o(2^k)$, eventually $G_k\le 2^{k-2}$, while the block $[2^k,2^{k+1})$ contains $2^{k-1}$ odd numbers.
• Does the theorem say anything directly about the prime-parity sequence? Not by itself. It classifies what can be concluded from a discrepancy-only estimate $A_\pi(x)=O(F(x))$. The actual prime sequence may have extra structure; the theorem rules out an entire class of proofs, but does not solve the Erdős problem unconditionally.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

This round proves a full abstract theorem that strictly strengthens Round 3:

• Round 2 gave a sufficient discrepancy criterion.
• Round 3 showed sharpness for the special family $F(x)=x/(\log\log x)^\alpha$.
• Round 4 gives the exact threshold for every monotone sublinear gauge $F$: the discrepancy-only implication \[ A(x)=O(F(x))\Longrightarrow \sum \frac{\varepsilon(n)}{n\log n}\text{ converges} \] holds for all sign sequences if and only if \[ \sum \frac{F(n)}{n^2\log n}<\infty. \]

For Erdős problem 15, this isolates the remaining issue sharply: any unconditional proof based only on a bound for \[ A_\pi(x)=\sum_{m\le x}(-1)^{\pi(m)} \] must beat the exact threshold above; anything weaker is provably insufficient in general.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 82%

9) REFERENCES

References

• [EP15] Erdős Problems website, Problem 15: Is it true that $\sum_{n=1}^\infty (-1)^n n/p_n$ converges? https://www.erdosproblems.com/latex/15.
• [Tao2024] T. Tao, The convergence of an alternating series of Erdős, assuming the Hardy--Littlewood prime tuples conjecture, Communications of the American Mathematical Society 4 $2024$, 80--96.
• [TaoArxiv] T. Tao, The convergence of an alternating series of Erdős, assuming the Hardy--Littlewood prime tuples conjecture, arXiv:2308.07205.

10) OUTPUT FORMAT

The Erdős problem number here is 15. This file is therefore exported as 15_solutions.tex, beginning directly with \section and with no preamble.