gpt pro 5.4 attempt on Math-644, version 3

Problem 15 --- Round 4

1) ROUND-4 OBJECTIVE

(C) Obstruction / correction. I continue the obstruction/correction track from Round 3. Round 3 reduced the original Erdős series \[ \sum_{n=1}^\infty (-1)^n\frac{n}{p_n} \] to the parity-weighted series \[ \sum_{m=3}^\infty \frac{(-1)^{\pi(m)}}{m\log m} \] and then to an exact Abel-type criterion involving the discrepancy \[ A(x):=\sum_{3\le m\le x}(-1)^{\pi(m)}. \] The natural Round 4 gap is therefore no longer the prime-indexed series itself, but the following sharper question:

Exactly when can a discrepancy bound of the form \[ A(x)=O\!\left(\frac{x}{L(x)}\right) \] force convergence of the weighted series?

The most promising Round 4 direction is to settle this question at the level of generic sign sequences. I will prove a sharp threshold theorem showing that, within the entire class of "discrepancy-only" arguments, the integral condition \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}<\infty \] is exactly the right scale: it is sufficient for convergence, and if it fails then one can explicitly build equidistributed sign sequences with discrepancy \[ O\!\left(\frac{x}{L(x)}\right) \] for which the analogous weighted series diverges.

This is a genuine Round 4 advance beyond Round 3: it does not solve the prime problem, but it rules out an entire class of proof strategies in a near-optimal way and gives the minimal corrected statement for what a discrepancy-only method could possibly prove.

2) Round-3 FOUNDATION USED

I use the following Round 3 results as fixed input.

1. Round 3 Theorem 6.2 (exact discrepancy criterion for the prime problem). Let \[ A(x):=\sum_{3\le m\le x}(-1)^{\pi(m)}. \] Then the original Erdős series \[ \sum_{n=1}^\infty (-1)^n\frac{n}{p_n} \] converges if and only if the parity series \[ \sum_{m=3}^\infty \frac{(-1)^{\pi(m)}}{m\log m} \] converges, and this in turn is equivalent to convergence of the discrepancy-weighted series/integral \[ \sum_{m=3}^\infty A(m)\left(\frac{1}{m\log m}-\frac{1}{(m+1)\log(m+1)}\right) \] or equivalently \[ \int_3^\infty A(t)\left(\frac{1}{t^2\log t}+\frac{1}{t^2\log^2 t}\right)dt. \]

2. Round 3 Corollary 6.3 (sufficient discrepancy criterion). If \[ A(x)=O\!\left(\frac{x}{L(x)}\right) \] for some scale $L(x)\to\infty$ with \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}<\infty, \] then the original Erdős series converges.

3. Round 3 Proposition 6.4 (one explicit model obstruction). There exists an equidistributed sign sequence with discrepancy \[ O\!\left(\frac{x}{\log\log x\,\log\log\log x}\right) \] for which the weighted series \[ \sum \frac{\sigma(m)}{m\log m} \] diverges.

I will not re-prove these Round 3 results. Round 4 begins from them and sharpens them.

3) NEW INSIGHT / TOOL (ROUND-4)

The new ingredient is a sharp threshold theorem for discrepancy-only methods.

1. I extend the Round 3 Abel criterion from the specific prime-parity sequence $(-1)^{\pi(m)}$ to arbitrary sign sequences $\sigma(m)\in\{\pm1\}$.

2. I prove a converse construction: whenever a scale $L(x)$ is too weak in the sense that \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}=\infty, \] there exists an explicit equidistributed sign sequence with discrepancy \[ O\!\left(\frac{x}{L(x)}\right) \] whose weighted series diverges.

3. Combining the two, I obtain the minimal corrected statement:

   For discrepancy-only proofs, the integral threshold \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}<\infty \] is optimal (up to mild monotonicity assumptions on $x/L(x)$).

This is strictly stronger than Round 3, which only supplied one sufficient criterion and one special counterexample scale.

4) ATTACK PLAN (ROUND-4)

After Round 3, the exact unresolved structural question was:

How sharp is the discrepancy criterion? In particular, which bounds on \[ A(x)=\sum_{3\le m\le x}(-1)^{\pi(m)} \] could ever suffice if one tries to prove convergence using only Abel summation and discrepancy estimates?

To close that gap, I will:

1. prove a generic Abel-summation theorem for arbitrary sign sequences;
2. prove a generic converse construction showing that the Round 3 sufficient condition is essentially optimal;
3. apply the result back to the prime problem via Round 3 Theorem 6.2, thereby ruling out an entire class of proof strategies.

This directly overcomes the Round 3 obstacle: instead of asking for a single ad hoc example, it identifies the exact integrability threshold that controls all discrepancy-only approaches.

5) WORK (ROUND-4)

5.1. A generic Abel identity for weighted sign series

Let \[ w_n:=\frac{1}{n\log n}\qquad (n\ge 3), \] and let $\sigma_n$ be any real sequence. Define its partial sums by \[ B(N):=\sum_{n=3}^N \sigma_n. \]

Lemma 7.1 (generic Abel identity). For every integer $N\ge 3$, \[ \sum_{n=3}^N \sigma_n w_n = B(N)w_N + \sum_{n=3}^{N-1} B(n)(w_n-w_{n+1}). \]

Proof. This is the standard summation-by-parts identity applied to $a_n=\sigma_n$ and $b_n=w_n$: \[ \sum_{n=3}^N a_nb_n = A(N)b_N + \sum_{n=3}^{N-1} A(n)(b_n-b_{n+1}), \] where $A(n)=\sum_{j=3}^n a_j=B(n)$. Substituting $a_n=\sigma_n$ and $b_n=w_n$ gives the claim. $\square$

Lemma 7.2 (kernel asymptotics). There exist absolute constants $0<c<C<\infty$ such that for all sufficiently large $n$, \[ \frac{c}{n^2\log n} \le w_n-w_{n+1} \le \frac{C}{n^2\log n}. \] More precisely, \[ w_n-w_{n+1} = \int_n^{n+1}\left(\frac{1}{t^2\log t}+\frac{1}{t^2\log^2 t}\right)dt. \]

Proof. The function $w(t)=1/(t\log t)$ is differentiable for $t>1$, and \[ -w'(t)=\frac{1}{t^2\log t}+\frac{1}{t^2\log^2 t}. \] Hence \[ w_n-w_{n+1} = \int_n^{n+1} -w'(t)\,dt = \int_n^{n+1}\left(\frac{1}{t^2\log t}+\frac{1}{t^2\log^2 t}\right)dt. \] Since on $[n,n+1]$ one has $t\asymp n$ and $\log t\asymp \log n$, the stated two-sided bound follows. $\square$

5.2. Generic sufficiency at a given discrepancy scale

The next statement extends Round 3 Corollary 6.3 from the prime-parity sequence to arbitrary sign sequences.

Proposition 7.3 (generic discrepancy-only sufficiency). Let $L:[3,\infty)\to(0,\infty)$ be eventually increasing with $L(x)\to\infty$. Let $\sigma_n\in\{\pm1\}$ and write \[ B(N):=\sum_{n=3}^N \sigma_n. \] Assume \[ B(N)=O\!\left(\frac{N}{L(N)}\right) \qquad (N\to\infty) \] and \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}<\infty. \] Then the series \[ \sum_{n=3}^\infty \frac{\sigma_n}{n\log n} \] converges.

Proof. By Lemma 7.1, \[ \sum_{n=3}^N \frac{\sigma_n}{n\log n} = \frac{B(N)}{N\log N} + \sum_{n=3}^{N-1} B(n)(w_n-w_{n+1}). \] The boundary term tends to zero because \[ \left|\frac{B(N)}{N\log N}\right| \ll \frac{1}{L(N)\log N} \to 0. \] Also, by Lemma 7.2 and the hypothesis on $B(n)$, \[ |B(n)|(w_n-w_{n+1}) \ll \frac{n/L(n)}{n^2\log n} = \frac{1}{n\log n\,L(n)}. \] Because $L$ is eventually increasing, the function $1/(x\log x\,L(x))$ is eventually decreasing, so the integral test yields \[ \sum_{n\ge 3}\frac{1}{n\log n\,L(n)}<\infty. \] Hence \[ \sum_{n\ge 3}|B(n)|(w_n-w_{n+1})<\infty, \] so the Abel-transformed series converges absolutely. Therefore \[ \sum_{n=3}^\infty \frac{\sigma_n}{n\log n} \] converges. $\square$

5.3. A sharp converse model construction

I now prove the key new theorem of Round 4. It subsumes the special Round 3 counterexample and shows that the preceding sufficiency criterion is essentially best possible for discrepancy-only arguments.

Theorem 7.4 (sharpness of discrepancy-only criteria for generic sign sequences). Let $L:[3,\infty)\to(0,\infty)$ be eventually increasing with $L(x)\to\infty$, and set \[ D(x):=\frac{x}{L(x)}. \] Assume in addition that $D$ is eventually nondecreasing. If \[ \int_3^\infty \frac{dt}{t\log t\,L(t)} = \infty, \] then there exists a sign sequence $\sigma_n\in\{\pm1\}$ such that:

1. (i) its discrepancy satisfies \[ \sum_{n=3}^N \sigma_n = 2D(N)+O(1) \qquad (N\to\infty), \] and in particular \[ \sum_{n=3}^N \sigma_n = O\!\left(\frac{N}{L(N)}\right); \]

2. (ii) the sequence is equidistributed (because $D(N)=o(N)$);

3. (iii) the weighted series diverges to $+\infty$: \[ \sum_{n=3}^\infty \frac{\sigma_n}{n\log n}=+\infty. \]

Proof. Because $L(x)\to\infty$ and $L$ is eventually increasing, there exists an odd integer $N_0\ge 3$ such that for all $x\ge N_0-2$, \[ L(x)>3. \] For such $x$ we have \[ D(x+2)-D(x) = \frac{x+2}{L(x+2)}-\frac{x}{L(x)} \le \frac{x+2}{L(x)}-\frac{x}{L(x)} = \frac{2}{L(x)}<1. \] Since $D$ is assumed eventually nondecreasing, it follows that for all sufficiently large $x$, \[ 0\le D(x+2)-D(x)<1. \tag{7.1} \]

Now define a set $Y$ of odd integers by \[ Y:=\left\{n\ge N_0: n\text{ odd and }\lfloor D(n)\rfloor>\lfloor D(n-2)\rfloor\right\}. \] Because of (7.1), each floor difference is either $0$ or $1$. Let \[ C(X):=\#\{n\in Y: n\le X\}. \] If $X^*$ denotes the largest odd integer $\le X$, then by telescoping, \[ C(X) = \sum_{\substack{N_0\le n\le X^*\\ n\text{ odd}}} \left(\lfloor D(n)\rfloor-\lfloor D(n-2)\rfloor\right) = \lfloor D(X^*)\rfloor-\lfloor D(N_0-2)\rfloor. \] Since $0\le D(X)-D(X^*)<1$ for large $X$, we obtain \[ C(X)=D(X)+O(1). \tag{7.2} \]

Define the sign sequence by flipping the alternating sequence on $Y$: \[ \sigma_n:=(-1)^n + 2\mathbf 1_Y(n). \] Because $Y$ consists only of odd integers, $\sigma_n\in\{\pm1\}$ for every $n$. Its discrepancy is \[ B(N):=\sum_{n=3}^N \sigma_n = \sum_{n=3}^N (-1)^n + 2C(N). \] The alternating partial sums are bounded, so by (7.2), \[ B(N)=2D(N)+O(1). \] This proves (i). Since $D(N)=N/L(N)=o(N)$, we also have \[ B(N)=o(N), \] which is exactly equidistribution of the two signs; thus (ii) holds.

It remains to prove (iii). Write \[ \sum_{n=3}^N \frac{\sigma_n}{n\log n} = \sum_{n=3}^N \frac{(-1)^n}{n\log n} + 2\sum_{\substack{n\in Y\\ n\le N}} \frac{1}{n\log n}. \tag{7.3} \] The first term converges by the alternating series test, so it suffices to show that \[ \sum_{n\in Y} \frac{1}{n\log n}=\infty. \] By summation by parts, \[ \sum_{\substack{n\in Y\\ n\le N}} w_n = C(N)w_N + \sum_{m=N_0}^{N-1} C(m)(w_m-w_{m+1}). \tag{7.4} \] The boundary term tends to zero because, by (7.2), \[ C(N)w_N \ll \frac{D(N)}{N\log N} = \frac{1}{L(N)\log N} \to 0. \] Also, again by (7.2), \[ \sum_{m=N_0}^{N-1} C(m)(w_m-w_{m+1}) = \sum_{m=N_0}^{N-1} D(m)(w_m-w_{m+1}) + O\!\left(\sum_{m=N_0}^{\infty}(w_m-w_{m+1})\right). \tag{7.5} \] The error term is finite because the latter series telescopes. Thus it is enough to prove that \[ \sum_{m\ge N_0} D(m)(w_m-w_{m+1})=\infty. \tag{7.6} \] By Lemma 7.2, \[ D(m)(w_m-w_{m+1}) \asymp \frac{D(m)}{m^2\log m} = \frac{1}{m\log m\,L(m)}. \] Because $L$ is eventually increasing, the latter sequence is eventually decreasing, so the integral test applies. Since by hypothesis \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}=\infty, \] we conclude that \[ \sum_{m\ge N_0} \frac{1}{m\log m\,L(m)}=\infty, \] and therefore (7.6) holds. Returning to (7.4) and (7.5), we obtain \[ \sum_{n\in Y} \frac{1}{n\log n}=\infty. \] Now (7.3) shows that \[ \sum_{n=3}^\infty \frac{\sigma_n}{n\log n}=+\infty, \] proving (iii). $\square$

5.4. Consequence for the original Erdős problem

The generic theorem above does not by itself solve the prime problem, but it gives a sharp obstruction for an entire proof strategy.

Corollary 7.5 (optimality of discrepancy-only methods for Problem 15). Let \[ A(x):=\sum_{3\le m\le x} (-1)^{\pi(m)}. \] Suppose one seeks to prove convergence of \[ \sum_{n=1}^\infty (-1)^n\frac{n}{p_n} \] using only a bound of the form \[ A(x)=O\!\left(\frac{x}{L(x)}\right) \] for some eventually increasing scale $L(x)\to\infty$ with $x/L(x)$ eventually nondecreasing. Then:

1. (a) if \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}<\infty, \] then such a discrepancy bound would suffice to prove convergence (by Round 3 Theorem 6.2 and Proposition 7.3);

2. (b) if \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}=\infty, \] then no argument based solely on that generic discrepancy scale can be sharp in full generality, because Theorem 7.4 produces equidistributed model sign sequences with the same discrepancy scale for which the analogous weighted series diverges.

Hence the threshold \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}<\infty \] is the minimal corrected statement for what a discrepancy-only strategy can hope to prove.

In particular, Round 3's special model obstruction at the scale \[ L(x)=\log\log x\,\log\log\log x \] is now seen to be a single instance of a much more general sharp theorem.

6) ADVERSARIAL VERIFICATION

I now try to break the new Round 4 arguments.

1. Does Theorem 7.4 really produce a set, not a multiset? Yes. Because $0\le D(x+2)-D(x)<1$ for large $x$, the floor difference \[ \lfloor D(n)\rfloor-\lfloor D(n-2)\rfloor \] is always either $0$ or $1$. So at each odd index we either place one point or none; no multiplicities occur.

2. Is the count $C(X)=D(X)+O(1)$ really valid? Yes. The exact telescoping identity gives \[ C(X)=\lfloor D(X^*)\rfloor-\lfloor D(N_0-2)\rfloor, \] where $X^*$ is the largest odd integer $\le X$. Since $D$ changes by less than $1$ across any interval of length $2$, replacing $X^*$ by $X$ changes $D$ by $O(1)$. Thus $C(X)=D(X)+O(1)$.

3. Could the divergence proof fail because the comparison with $1/(m\log mL(m))$ is too weak? No. Lemma 7.2 gives the two-sided estimate \[ D(m)(w_m-w_{m+1})\asymp \frac{1}{m\log m\,L(m)}. \] Since $L$ is eventually increasing, the comparison sequence is eventually decreasing, so the integral test is legitimate.

4. Is equidistribution automatic? Yes. The discrepancy of the constructed sign sequence is \[ B(N)=2D(N)+O(1)=O\!\left(\frac{N}{L(N)}\right), \] and because $L(N)\to\infty$, this is $o(N)$. That is exactly the statement that the two signs occur with asymptotic density $1/2$.

5. Did Round 4 improperly discard Round 3's prime-specific structure? No. Round 4 does not replace the prime problem by an unrelated model; rather, it uses Round 3 Theorem 6.2 to identify the prime problem with one special weighted sign sequence, and then proves a general theorem showing what can and cannot follow from discrepancy information alone. This is precisely the kind of obstruction analysis requested in path (C).

6. What mild regularity is still assumed? The converse model theorem assumes that $L$ is eventually increasing and that $D(x)=x/L(x)$ is eventually nondecreasing. This covers the natural scales that arise in practice (for instance powers of $\log\log x$, powers of $\log\log\log x$, and their products). I do not claim here to have removed every monotonicity assumption in complete generality.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

Round 4 proves a substantially sharper obstruction theorem than Round 3.

1. It generalizes the Round 3 Abel criterion from the prime-parity sequence to arbitrary sign sequences.

2. It proves a sharp converse construction: for every eventually increasing scale $L(x)$ with $x/L(x)$ eventually nondecreasing and \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}=\infty, \] there exists an equidistributed sign sequence with discrepancy \[ O\!\left(\frac{x}{L(x)}\right) \] whose weighted series \[ \sum \frac{\sigma_n}{n\log n} \] diverges.

3. Consequently, for Problem 15, any attempt to prove convergence using only a discrepancy estimate of the form \[ A(x)=O\!\left(\frac{x}{L(x)}\right) \] is optimal exactly at the integral threshold \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}<\infty. \] This is the minimal corrected statement for the discrepancy-only strategy.

What remains open is the genuinely prime-specific question: does the actual discrepancy \[ A(x)=\sum_{3\le m\le x}(-1)^{\pi(m)} \] exhibit enough cancellation, beyond generic discrepancy considerations, to make the Abel kernel converge unconditionally?

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 86%

9) REFERENCES

1. {[EP }] Erdős Problems website, Problem 15: the entry for \[ \sum_{n=1}^\infty (-1)^n\frac{n}{p_n} \] and related conjectures on reciprocal prime-gap series.

2. {[Ta23 }] T. Tao, The convergence of an alternating series of Erdős, assuming the Hardy--Littlewood prime tuples conjecture, arXiv:2308.07205.

10) OUTPUT FORMAT

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