Erdős Problem #15

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

This text is written as a drop-in LaTeX fragment with no preamble and no \begin{document}.

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).

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).

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).

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.

Step 2: divergence of the weighted series.

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).

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.

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).

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).

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.

Problem 15 --- Round 3

1) ROUND-3 OBJECTIVE

(C) Obstruction / correction. I continue the obstruction/correction track from Round 2. Using the Round 2 equivalence between \[ \sum_{n=1}^\infty (-1)^n \frac{n}{p_n} \quad\text{and}\quad \sum_{m=3}^\infty \frac{(-1)^{\pi(m)}}{m\log m}, \] the natural next gap is no longer the prime-indexed series itself, but rather the exact amount of discrepancy cancellation needed for the parity sequence \[ \varepsilon_m := (-1)^{\pi(m)}. \]

The most promising Round 3 direction is therefore:

• derive an exact necessary-and-sufficient criterion for convergence in terms of the discrepancy \[ A(x):=\sum_{3\le m\le x}\varepsilon_m, \] and
• show that the Round 2 sufficient condition is essentially sharp in the model setting by constructing an explicit equidistributed sign sequence with discrepancy as small as \[ O\!\left(\frac{x}{\log\log x\,\log\log\log x}\right) \] for which the weighted series \[ \sum \frac{\varepsilon_m}{m\log m} \] still diverges.

This strictly advances Round 2 in two ways: it replaces a merely sufficient discrepancy condition by an exact one, and it rules out an entire stronger class of discrepancy-only proof strategies.

2) Round-2 FOUNDATION USED

I use the following Round 2 results as fixed input.

1. Round 2 Theorem 5.7 (Said--Tao equivalence). There exists a constant $C\in\mathbb R$ such that \[ \sum_{n\le x} (-1)^n \frac{n}{p_n} = \frac12 \sum_{3\le m\le x\log x} \frac{(-1)^{\pi(m)}}{m\log m} + C + o(1). \] Hence the original Erdős series converges if and only if \[ \sum_{m=3}^\infty \frac{(-1)^{\pi(m)}}{m\log m} \] converges.

2. Round 2 Proposition 5.8 (one sufficient discrepancy criterion). If \[ A(x)=O\!\left(\frac{x}{(\log\log x)^{1+\varepsilon}}\right) \] for some $\varepsilon>0$, then the parity series, and hence the original Erdős series, converges.

3. Round 2 Proposition 5.9 (equidistribution alone is insufficient). There exists an equidistributed sign sequence for which \[ \sum \frac{\varepsilon_m}{m\log m} \] diverges.

I will not re-prove these results; Round 3 begins from them.

3) NEW INSIGHT / TOOL (ROUND-3)

The new ingredients are the following.

1. Exact Abel criterion. The parity series is convergent if and only if a certain discrepancy-weighted series \[ \sum_{m\ge 3} A(m)\left(\frac1{m\log m}-\frac1{(m+1)\log(m+1)}\right) \] converges. Equivalently, it is convergent if and only if the improper integral \[ \int_3^\infty A(t)\left(\frac{1}{t^2\log t}+\frac{1}{t^2\log^2 t}\right)\,dt \] converges.

2. A strictly stronger general sufficient criterion. If $L(x)\to\infty$ and \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}<\infty, \] then \[ A(x)=O\!\left(\frac{x}{L(x)}\right) \] already implies convergence. This sharpens Round 2 Proposition 5.8.

3. Near-sharp model obstruction. I construct an explicit equidistributed sign sequence with discrepancy \[ O\!\left(\frac{x}{\log\log x\,\log\log\log x}\right) \] whose $1/(m\log m)$-weighted series diverges. Thus one cannot hope to prove convergence from any discrepancy estimate of that scale alone.

4) ATTACK PLAN (ROUND-3)

After Round 2, the remaining gap is:

What quantitative control on \[ A(x)=\sum_{3\le m\le x} (-1)^{\pi(m)} \] is actually equivalent to convergence of \[ \sum_{m=3}^\infty \frac{(-1)^{\pi(m)}}{m\log m}\,? \]

The Round 2 result only gave a sufficient condition. To advance beyond that, I will do two things.

1. Starting from Round 2 Theorem 5.7, I will derive an exact summation-by-parts identity for the parity series, yielding a necessary-and-sufficient convergence criterion in terms of $A(x)$.

2. I will then build an explicit model sign sequence showing that even the much stronger discrepancy bound \[ A(x)=O\!\left(\frac{x}{\log\log x\,\log\log\log x}\right) \] does not force convergence in general. This strengthens the Round 2 obstruction, which only ruled out plain equidistribution.

This attacks the exact Round 2 gap: it identifies the correct kernel against which the discrepancy must cancel, and it proves that a whole class of weaker discrepancy-only strategies cannot possibly settle the original problem.

5) WORK (ROUND-3)

5.1. Exact Abel criterion for the parity series

Set \[ \varepsilon_m := (-1)^{\pi(m)},\qquad T(N):=\sum_{m=3}^N \frac{\varepsilon_m}{m\log m},\qquad A(N):=\sum_{m=3}^N \varepsilon_m \] for integers $N\ge 3$.

By Round 2 Theorem 5.7, the original Erdős series converges if and only if $T(N)$ converges as $N\to\infty$.

Lemma 6.1 (discrete Abel identity). For every integer $N\ge 3$, \[ T(N) = \frac{A(N)}{N\log N} + \sum_{m=3}^{N-1} A(m)\left( \frac{1}{m\log m} - \frac{1}{(m+1)\log(m+1)} \right). \]

Proof. Apply the standard summation-by-parts identity \[ \sum_{m=3}^N a_m b_m = A(N)b_N + \sum_{m=3}^{N-1} A(m)(b_m-b_{m+1}), \] where $A(m):=\sum_{j=3}^m a_j$, with \[ a_m=\varepsilon_m,\qquad b_m=\frac{1}{m\log m}. \] This gives the claimed identity immediately. $\square$

Theorem 6.2 (exact discrepancy criterion). Let \[ A(N)=\sum_{m=3}^N (-1)^{\pi(m)}. \] Then the following are equivalent:

1. (i) the series \[ \sum_{n=1}^\infty (-1)^n \frac{n}{p_n} \] converges;

2. (ii) the series \[ \sum_{m=3}^\infty \frac{(-1)^{\pi(m)}}{m\log m} \] converges;

3. (iii) the discrepancy-weighted series \[ \sum_{m=3}^\infty A(m)\left( \frac{1}{m\log m} - \frac{1}{(m+1)\log(m+1)} \right) \] converges.

Moreover, if $A(t):=A(\lfloor t\rfloor)$ for real $t\ge 3$, then these are also equivalent to convergence of the improper integral \[ \int_3^\infty A(t)\left(\frac{1}{t^2\log t}+\frac{1}{t^2\log^2 t}\right)\,dt. \]

Proof. The equivalence of (i) and (ii) is exactly Round 2 Theorem 5.7.

By Lemma 6.1, \[ T(N) = \frac{A(N)}{N\log N} + \sum_{m=3}^{N-1} A(m)\left( \frac{1}{m\log m} - \frac{1}{(m+1)\log(m+1)} \right). \] Since $|A(N)|\le N-2$, we have \[ \left|\frac{A(N)}{N\log N}\right| \le \frac{N-2}{N\log N} \le \frac{1}{\log N} \to 0. \] Hence $T(N)$ converges if and only if the partial sums of \[ \sum_{m=3}^\infty A(m)\left( \frac{1}{m\log m} - \frac{1}{(m+1)\log(m+1)} \right) \] converge. This proves the equivalence of (ii) and (iii), hence also of (i)--(iii).

For the integral form, note that \[ \frac{d}{dt}\!\left(\frac{1}{t\log t}\right) = -\frac{1}{t^2\log t}-\frac{1}{t^2\log^2 t}. \] Therefore, for every integer $m\ge 3$, \[ \frac{1}{m\log m} - \frac{1}{(m+1)\log(m+1)} = \int_m^{m+1} \left( \frac{1}{t^2\log t} + \frac{1}{t^2\log^2 t} \right)\,dt. \] Because $A(t)=A(m)$ on $t\in[m,m+1)$, we obtain for every integer $N\ge 4$, \[ \sum_{m=3}^{N-1} A(m)\left( \frac{1}{m\log m} - \frac{1}{(m+1)\log(m+1)} \right) = \int_3^N A(t)\left( \frac{1}{t^2\log t} + \frac{1}{t^2\log^2 t} \right)\,dt. \] Also, \[ \int_N^{N+1} \left| A(t)\left( \frac{1}{t^2\log t} + \frac{1}{t^2\log^2 t} \right) \right|\,dt \ll N\int_N^{N+1}\frac{dt}{t^2\log t} \ll \frac{1}{N\log N}\to 0, \] so convergence of the integrals along integer endpoints $N$ is equivalent to convergence of the full improper integral. Thus the discrepancy series converges if and only if the improper integral converges. $\square$

This is an exact criterion; unlike the Round 2 Proposition 5.8, it is not merely sufficient.

5.2. A stronger general sufficient criterion

Theorem 6.2 immediately yields the following strengthening of Round 2 Proposition 5.8.

Corollary 6.3 (general discrepancy criterion). Let $L:[3,\infty)\to (0,\infty)$ be eventually increasing with $L(x)\to\infty$. Assume \[ A(x)=O\!\left(\frac{x}{L(x)}\right) \] and \[ \int_3^\infty \frac{dt}{t\log t\,L(t)}<\infty. \] Then \[ \sum_{n=1}^\infty (-1)^n\frac{n}{p_n} \] converges.

Proof. By Theorem 6.2, it suffices to show that \[ \int_3^\infty A(t)\left(\frac{1}{t^2\log t}+\frac{1}{t^2\log^2 t}\right)\,dt \] converges. From the hypothesis, \[ |A(t)| \ll \frac{t}{L(t)}. \] Hence \[ \left| A(t)\left(\frac{1}{t^2\log t}+\frac{1}{t^2\log^2 t}\right) \right| \ll \frac{1}{t\log t\,L(t)}. \] The right-hand side is integrable by assumption, so the integral converges absolutely. Therefore the original Erdős series converges. $\square$

Remark. Round 2 Proposition 5.8 corresponds to the special choice \[ L(t)=(\log\log t)^{1+\varepsilon}. \] But Corollary 6.3 is strictly stronger. For instance, it already gives convergence from the much weaker-looking bound \[ A(x)=O\!\left( \frac{x}{\log\log x\,(\log\log\log x)^{1+\varepsilon}} \right), \] because \[ \int_3^\infty \frac{dt}{t\log t\,\log\log t\,(\log\log\log t)^{1+\varepsilon}} <\infty. \]

5.3. Near-sharp model obstruction

I now strengthen the Round 2 obstruction.

Proposition 6.4 (near-sharp model counterexample). There exists an equidistributed sign sequence $\sigma(m)\in\{\pm 1\}$ such that, with \[ B(x):=\sum_{3\le m\le x}\sigma(m), \] one has \[ B(x)=O\!\left(\frac{x}{\log\log x\,\log\log\log x}\right), \] but nevertheless \[ \sum_{m=3}^\infty \frac{\sigma(m)}{m\log m} \] diverges (in fact, diverges to $+\infty$).

Proof. Let \[ L_k:=\log\log k\,\log\log\log k \] for $k$ sufficiently large, and define \[ n_k:=2\lfloor kL_k\rfloor+1. \] For $k$ large enough, $L_k>1$, so $n_k$ is odd and strictly increasing. Let \[ Y:=\{n_k: k\ge k_0\} \] for some sufficiently large initial index $k_0$, and define \[ \sigma(m):= \begin{cases} 1, & m\in Y,\\ (-1)^m, & m\notin Y. \end{cases} \]

Thus $\sigma$ agrees with the alternating sequence $(-1)^m$, except that at the selected odd integers $m\in Y$ we flip the value from $-1$ to $+1$.

Step 1: discrepancy bound. Let \[ Y(x):=\#\{m\in Y: m\le x\}. \] I claim that \[ Y(x)=O\!\left(\frac{x}{\log\log x\,\log\log\log x}\right). \]

Indeed, fix $x$ large and suppose $n_k\le x$. Write \[ M(x):=\log\log x\,\log\log\log x. \] If \[ k>\frac{8x}{M(x)}, \] then for $x$ large enough one also has $k>x^{1/2}$. Hence \[ \log\log k \ge \frac12 \log\log x, \qquad \log\log\log k \ge \frac12 \log\log\log x, \] so \[ L_k=\log\log k\,\log\log\log k \ge \frac14 M(x). \] Therefore \[ n_k \ge kL_k > \frac{8x}{M(x)}\cdot \frac14 M(x) = 2x, \] contradicting $n_k\le x$. Thus necessarily \[ k\le \frac{8x}{M(x)}, \] and so \[ Y(x)\le \frac{8x}{\log\log x\,\log\log\log x} \] for all sufficiently large $x$, proving the claim.

Now let \[ B_0(x):=\sum_{3\le m\le x} (-1)^m. \] Then $B_0(x)=O(1)$. Every $m\in Y$ is odd, so replacing $(-1)^m=-1$ by $+1$ changes the partial sum by $+2$. Hence \[ B(x)=B_0(x)+2Y(x), \] and therefore \[ B(x)=O(1)+O\!\left(\frac{x}{\log\log x\,\log\log\log x}\right) = O\!\left(\frac{x}{\log\log x\,\log\log\log x}\right). \] In particular $B(x)=o(x)$, so $\sigma$ is equidistributed.

Step 2: divergence of the weighted series. We have the exact decomposition \[ \sum_{m=3}^\infty \frac{\sigma(m)}{m\log m} = \sum_{m=3}^\infty \frac{(-1)^m}{m\log m} + \sum_{m\in Y} \frac{\sigma(m)-(-1)^m}{m\log m}. \] Since every $m\in Y$ is odd, $(-1)^m=-1$ and $\sigma(m)=1$, so \[ \sigma(m)-(-1)^m=2. \] Therefore \[ \sum_{m=3}^\infty \frac{\sigma(m)}{m\log m} = \sum_{m=3}^\infty \frac{(-1)^m}{m\log m} + 2\sum_{m\in Y}\frac{1}{m\log m}. \] The first series converges by the alternating series test. Thus it remains to prove \[ \sum_{m\in Y}\frac{1}{m\log m}=\infty. \]

Since $m=n_k$ for $m\in Y$, it is enough to prove \[ \sum_{k\ge k_0} \frac{1}{n_k\log n_k}=\infty. \] For $k$ sufficiently large, \[ n_k = 2\lfloor kL_k\rfloor + 1 \le 3kL_k. \] Also $L_k=o(k)$, so $n_k\le k^2$ for large $k$, and hence \[ \log n_k \le 2\log k. \] Thus \[ \frac{1}{n_k\log n_k} \ge \frac{c}{k\log k\,\log\log k\,\log\log\log k} \] for some absolute constant $c>0$ and all large $k$.

Now \[ \sum_{k\ge k_0} \frac{1}{k\log k\,\log\log k\,\log\log\log k} \] diverges by the integral test: with $u=\log\log k$, one has $du = dk/(k\log k)$, and the corresponding integral becomes \[ \int^\infty \frac{du}{u\log u}, \] which diverges. Therefore \[ \sum_{k\ge k_0} \frac{1}{n_k\log n_k}=\infty, \] so \[ \sum_{m\in Y}\frac{1}{m\log m}=\infty. \] It follows that \[ \sum_{m=3}^\infty \frac{\sigma(m)}{m\log m}=+\infty. \] This proves the proposition. $\square$

Corollary 6.5 (strategy barrier). No theorem of the form \[ B(x)=O\!\left(\frac{x}{\log\log x\,\log\log\log x}\right) \quad\Longrightarrow\quad \sum_{m=3}^\infty \frac{\sigma(m)}{m\log m}\ \text{converges} \] can hold for arbitrary sign sequences $\sigma(m)\in\{\pm1\}$.

Thus any unconditional proof of convergence for the prime-parity sequence $\sigma(m)=(-1)^{\pi(m)}$ must use either:

• genuinely stronger cancellation than that scale, or
• some additional prime-specific structure beyond a bare discrepancy estimate of that order.

This is a mathematically sharper obstruction than the Round 2 statement that "equidistribution alone is insufficient."

6) ADVERSARIAL VERIFICATION

I now try to break the new arguments.

1. Did Theorem 6.2 secretly assume convergence of the discrepancy series? No. The proof is a direct finite summation-by-parts identity, valid for each $N$, followed by the trivial bound \[ \left|\frac{A(N)}{N\log N}\right|\le \frac1{\log N}\to 0. \] No prior convergence assumption is used.

2. Is the integral criterion exact or merely asymptotic? It is exact. The identity \[ \frac{1}{m\log m}-\frac{1}{(m+1)\log(m+1)} = \int_m^{m+1}\left(\frac{1}{t^2\log t}+\frac{1}{t^2\log^2 t}\right)\,dt \] holds exactly, and $A(t)=A(m)$ on $[m,m+1)$. So the discrepancy series and the integral are term-by-term identical after grouping by unit intervals.

3. Could Proposition 6.4 fail because the set $Y$ is not sparse enough? No. The counting argument shows \[ Y(x)\ll \frac{x}{\log\log x\,\log\log\log x}, \] which is $o(x)$. Hence the modified sequence is genuinely equidistributed.

4. Could the lower bound on $1/(n_k\log n_k)$ fail? No. For large $k$, \[ n_k\le 3k\log\log k\,\log\log\log k \] and also $n_k\le k^2$, so $\log n_k\le 2\log k$. This gives the required lower bound \[ \frac{1}{n_k\log n_k} \gg \frac{1}{k\log k\,\log\log k\,\log\log\log k}. \]

5. Does Proposition 6.4 contradict Round 2 Proposition 5.8? No. Round 2 Proposition 5.8 required the much stronger bound \[ A(x)=O\!\left(\frac{x}{(\log\log x)^{1+\varepsilon}}\right), \] while Proposition 6.4 only produces \[ B(x)=O\!\left(\frac{x}{\log\log x\,\log\log\log x}\right). \] Since \[ \frac{x}{\log\log x\,\log\log\log x} \gg \frac{x}{(\log\log x)^{1+\varepsilon}} \] for every fixed $\varepsilon>0$, there is no inconsistency.

6. Interaction with Round 2. Round 2 isolated the prime-parity discrepancy $A(x)$ as the key quantity. Round 3 now identifies the exact kernel against which $A(x)$ must cancel, and shows that the discrepancy scale needed for a general convergence theorem is very close to the natural integrability threshold. This is a genuine strengthening, not a reformulation.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

Round 3 makes two substantive advances beyond Round 2:

1. It upgrades the Round 2 sufficient criterion to an exact necessary-and-sufficient criterion: the original Erdős series converges if and only if \[ \int_3^\infty A(t)\left(\frac{1}{t^2\log t}+\frac{1}{t^2\log^2 t}\right)\,dt \] converges, where \[ A(t)=\sum_{3\le m\le t} (-1)^{\pi(m)}. \]

2. It proves a near-sharp model obstruction: there exists an equidistributed sign sequence with discrepancy \[ O\!\left(\frac{x}{\log\log x\,\log\log\log x}\right) \] whose $1/(m\log m)$-weighted series diverges. Therefore, discrepancy-only arguments at that scale cannot settle the prime problem.

What remains open is still the prime-specific question: can one prove, unconditionally, that the actual discrepancy \[ A(x)=\sum_{3\le m\le x} (-1)^{\pi(m)} \] has enough cancellation against the kernel from Theorem 6.2 to make the integral converge?

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 80%

9) REFERENCES

1. T. Tao, The convergence of an alternating series of Erdős, assuming the Hardy--Littlewood prime tuples conjecture, arXiv:2308.07205.

2. T. Tao, The convergence of an alternating series of Erdős, assuming the Hardy--Littlewood prime tuples conjecture, blog post, 14 Aug. 2023.

10) OUTPUT FORMAT

This text is written as a drop-in LaTeX fragment with no preamble and no \begin{document}.