gpt pro 5.2 attempt on Math-644, version 2

1) ROUND-2 OBJECTIVE

(C) Obstruction / correction. Given the Round‑1 status (UNRESOLVED) and the known state of the art (Tao proves convergence only under a strong Hardy–Littlewood prime tuples hypothesis), the most productive Round‑2 move is to isolate the exact analytic obstruction and give a minimal “corrected” criterion that would imply convergence. Concretely, I will prove a precise equivalence between the Erdős series \[ \sum_{n\ge1}(-1)^n\frac{n}{p_n} \] and a series governed by the parity of the prime counting function $(-1)^{\pi(m)}$. This reveals that mere equidistribution of parity is far too weak, and I will give an explicit counterexample in a model setting (an equidistributed ${ \pm1}$-sequence whose $1/(n\log n)$-weighted sum diverges).

This is a genuine “gap closure”: it converts the original prime-indexed question into an integer-indexed parity-discrepancy problem and rules out a tempting but insufficient strategy (“equidistribution $\Rightarrow$ convergence”).

(Background: this equivalence is attributed to an observation of Said and used by Tao; see Tao’s blog/paper and the original MO discussion. ([What's new][1]))

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2) ROUND-1 FOUNDATION USED

I will rely on the following Round‑1 items without re-proving:

1. Lemma 15.1 (increment formula): for $a_n=n/p_n$ and $g_n=p_{n+1}-p_n$, \[ a_{n+1}-a_n=\frac{p_n-n g_n}{p_n(p_n+g_n)}=\frac{p_n-n g_n}{p_n p_{n+1}}. \] 2. Pairing idea / summation-by-parts viewpoint: the alternating sum is controlled by differences $a_{n+1}-a_n$ on odd (n) (Round‑1 attack plan). 3. Numerics (used only as a sanity check): e.g. $S(10^6)\approx -0.0198592163$.

I will also use standard known results (outside Round‑1): prime number theorem-level bounds for $p_n$ $e.g. (p_n\asymp n\log n)$, and basic calculus estimates.

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3) NEW INSIGHT / TOOL (ROUND-2)

New Tool 1: Exact reduction to a prime-gap weighted sum

A clean “shift-and-average” identity converts the original series into \[ \frac12\sum_{n}(-1)^n\frac{n(p_{n+1}-p_n)}{p_n p_{n+1}} \quad+\quad\text{(a convergent correction)}. \] This makes the enemy (odd/even bias in gaps) explicit.

New Tool 2: Said–Tao equivalence to a parity-of-$\pi$ series

I will prove: \[ \sum_{n\le x}(-1)^n\frac{n}{p_n} ================================ \frac12\sum_{m\le x\log x}\frac{(-1)^{\pi(m)}}{m\log m} ;+;C;+;o(1), \] hence convergence of Erdős’ series is equivalent to convergence of \[ \sum_{m\ge 2}\frac{(-1)^{\pi(m)}}{m\log m}. \] This is precisely the “fatal obstruction”: the problem becomes one of quantitative cancellation in $(-1)^{\pi(m)}$, not just density $1/2$. ([What's new][1])

New Tool 3: Equidistribution alone is not enough

I give an explicit construction of an equidistributed sign sequence $\varepsilon(m)\in{\pm1}$ for which $\sum \varepsilon(m)/(m\log m)$ diverges. Thus, even if $\pi(m)\bmod 2$ were equidistributed, that would not settle convergence.

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4) ATTACK PLAN (ROUND-2)

Round‑1 gap: No unconditional control of $\sum (-1)^n n/p_n$.

What I will prove now:

1. A rigorous identity reducing the Erdős partial sums to a signed prime-gap sum plus a convergent correction (new lemma). 2. A rigorous equivalence (up to constant $+o(1)$) between Erdős partial sums and the parity series $\sum (-1)^{\pi(m)}/(m\log m)$ (new theorem). 3. A sharp “strategy barrier”: show that equidistribution of parity is insufficient by an explicit equidistributed-divergent construction. 4. Provide a minimal corrected sufficient condition: a quantitative discrepancy bound on $A(x):=\sum_{m\le x}(-1)^{\pi(m)}$ that would imply convergence.

This overcomes Round‑1’s obstacle by turning the prime-indexed alternation into a much cleaner integer-indexed cancellation problem.

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5) WORK (ROUND-2)

5.1 Notation

* $p_n$ = (n)-th prime; $g_n:=p_{n+1}-p_n$. * $\pi(m)$ = number of primes $\le m$. * Erdős partial sums: \[ S(x):=\sum_{n\le x}(-1)^n\frac{n}{p_n}. \] * Parity-series partial sums: \[ T(y):=\sum_{2\le m\le y}\frac{(-1)^{\pi(m)}}{m\log m}. \]

We use $a_n:=n/p_n$, so the term is $(-1)^n a_n$.

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5.2 Lemma: shift-and-average reduces to gaps

Lemma 5.1 (shift-and-average identity). As $x\to\infty$, \[ S(x)= -\frac14 + \frac12\sum_{n\le x}\left[(-1)^n\frac{n}{p_n}+(-1)^{n+1}\frac{n+1}{p_{n+1}}\right] ;+; o(1). \]

Proof. Define \[ S_{\mathrm{sh}}(x):=\sum_{n\le x}(-1)^{n+1}\frac{n+1}{p_{n+1}}. \] Change index $m=n+1$: \[ S_{\mathrm{sh}}(x)=\sum_{m=2}^{x+1}(-1)^m\frac{m}{p_m}=S(x+1)+\frac12, \] because $S(x+1)=(-1)^1\frac1{p_1}+\sum_{m=2}^{x+1}(-1)^m\frac{m}{p_m}=-\frac12+\sum_{m=2}^{x+1}\cdots$.

Thus \[ S_{\mathrm{sh}}(x)=S(x)+(-1)^{x+1}\frac{x+1}{p_{x+1}}+\frac12. \] Since $p_{x+1}\sim (x+1)\log(x+1)$, we have $(x+1)/p_{x+1}\to 0$, so \[ S_{\mathrm{sh}}(x)=S(x)+\frac12+o(1). \] Averaging (S(x)) and $S_{\mathrm{sh}}(x)-\frac12$ yields the claimed formula: \[ S(x)=\frac12\left(S(x)+S_{\mathrm{sh}}(x)-\frac12\right)+o(1) = -\frac14+\frac12\sum_{n\le x}\left[(-1)^n\frac{n}{p_n}+(-1)^{n+1}\frac{n+1}{p_{n+1}}\right]+o(1). \] ∎

Now expand the bracket.

Lemma 5.2 (gap decomposition inside the bracket). For every $n\ge1$, \[ (-1)^n\frac{n}{p_n}+(-1)^{n+1}\frac{n+1}{p_{n+1}} ================================================= (-1)^n\frac{n(p_{n+1}-p_n)}{p_n p_{n+1}} ;-; (-1)^n\frac{1}{p_{n+1}}. \]

Proof (algebra). \[ (-1)^{n+1}\frac{n+1}{p_{n+1}} =(-1)^n\left(-\frac{n}{p_{n+1}}-\frac{1}{p_{n+1}}\right), \] so the bracket equals \[ (-1)^n\left(\frac{n}{p_n}-\frac{n}{p_{n+1}}-\frac1{p_{n+1}}\right) ================================================================= (-1)^n\left(\frac{n(p_{n+1}-p_n)}{p_n p_{n+1}}-\frac1{p_{n+1}}\right). \] ∎

Combining Lemma 5.1 and 5.2:

Corollary 5.3 (Erdős sums = gap-weighted sum + convergent correction). There exists a constant $C_1$ such that \[ S(x) ==== \frac12\sum_{n\le x}(-1)^n\frac{n g_n}{p_n p_{n+1}} ;+;C_1;+;o(1). \] Indeed, $\sum_{n}(-1)^n/p_{n+1}$ converges by the alternating series test since $p_{n+1}$ increases.

This already refines Round‑1’s “pairing” obstacle: the entire question is now equivalent to understanding \[ \sum_{n\ge1}(-1)^n\frac{n g_n}{p_n p_{n+1}}. \]

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5.3 Main theorem: equivalence to $\sum (-1)^{\pi(m)}/(m\log m)$

This is the key new gap-closure.

Step A: Decompose (T(y)) into prime-gap blocks

For each (n), $\pi(m)=n$ for all integers $m\in[p_n,;p_{n+1}-1]$. Hence \[ (-1)^{\pi(m)}=(-1)^n \quad \text{for } p_n\le m<p_{n+1}. \] Therefore, for any $x\ge1$, \[ T(p_{x+1}-1)=\sum_{n\le x}(-1)^n\sum_{m=p_n}^{p_{n+1}-1}\frac{1}{m\log m}. \tag{5.1} \]

Step B: Approximate each block by $n g_n/(p_n p_{n+1})$ with absolutely summable error

Let $f(t)=\frac{1}{t\log t}$ for $t\ge 3$. Then (f) is positive and decreasing.

Define \[ \Sigma_n:=\sum_{m=p_n}^{p_{n+1}-1} f(m) =\sum_{m=p_n}^{p_{n+1}-1}\frac{1}{m\log m}. \]

Lemma 5.4 (discrete sum vs integral). \[ 0\le \Sigma_n - \int_{p_n}^{p_{n+1}} f(t),dt \le f(p_n)=\frac{1}{p_n\log p_n}. \]

Proof. Since (f) decreases, \[ \int_{m}^{m+1} f(t),dt \le f(m) \le \int_{m-1}^{m} f(t),dt \] and summing over $m=p_n,\dots,p_{n+1}-1$ gives the bound. ∎

Because $p_n\asymp n\log n$, we have $1/(p_n\log p_n)\asymp 1/(n\log^2 n)$, and thus \[ \sum_{n\ge3}\frac{1}{p_n\log p_n}<\infty. \tag{5.2} \] So the discrete-vs-integral error is absolutely summable.

Now handle the integral: \[ I_n:=\int_{p_n}^{p_{n+1}} \frac{dt}{t\log t}. \] Since (f) is continuous, by the mean value theorem for integrals there exists $\xi_n\in[p_n,p_{n+1}]$ such that \[ I_n=(p_{n+1}-p_n)\frac{1}{\xi_n\log \xi_n} =g_n\frac{1}{\xi_n\log \xi_n}. \tag{5.3} \]

We compare $\frac{1}{\xi_n\log \xi_n}$ to $\frac{n}{p_n p_{n+1}}$.

Use the standard PNT-level asymptotic (uniform in (n)) that \[ p_n = n\log n\left(1+O!\left(\frac{\log\log n}{\log n}\right)\right), \quad \log p_n = \log n + O(\log\log n), \tag{5.4} \] and the same for $p_{n+1}$, and also for any $\xi_n\in[p_n,p_{n+1}]$ (since $\xi_n\asymp n\log n$ as well).

From $5.4$ one obtains the expansions \[ \frac{1}{\xi_n\log \xi_n} ========================= \frac{1}{n\log^2 n} \left(1+O!\left(\frac{\log\log n}{\log n}\right)\right), \tag{5.5} \] and similarly \[ \frac{n}{p_n p_{n+1}} ===================== \frac{1}{n\log^2 n} \left(1+O!\left(\frac{\log\log n}{\log n}\right)\right). \tag{5.6} \] Subtracting gives \[ \frac{1}{\xi_n\log \xi_n}-\frac{n}{p_n p_{n+1}} =============================================== O!\left(\frac{\log\log n}{n\log^3 n}\right). \tag{5.7} \] Multiplying by $g_n$ and using $5.3$: \[ I_n-\frac{n g_n}{p_n p_{n+1}} ============================= g_n\cdot O!\left(\frac{\log\log n}{n\log^3 n}\right). \tag{5.8} \]

So it remains to show the absolute convergence of \[ \sum_{n\ge3} g_n\frac{\log\log n}{n\log^3 n}. \tag{5.9} \]

Lemma 5.5 (summability of the gap-weighted error). The series in $5.9$ converges.

Proof. Let \[ b_n:=\frac{\log\log n}{n\log^3 n}, \] which is positive and decreasing for (n) large. Use summation by parts with $g_n=p_{n+1}-p_n$: \[ \sum_{n=3}^{N} b_n g_n = \sum_{n=3}^{N} b_n(p_{n+1}-p_n) = b_N p_{N+1}-b_3 p_3 + \sum_{n=3}^{N-1} p_{n+1}(b_n-b_{n+1}). \tag{5.10} \] Now $p_{n+1}\ll n\log n$ and $b_n-b_{n+1}\ll \frac{\log\log n}{n^2\log^3 n}$, hence \[ p_{n+1}(b_n-b_{n+1}) \ll (n\log n)\cdot \frac{\log\log n}{n^2\log^3 n} ============================================= \frac{\log\log n}{n\log^2 n}, \] and $\sum_{n\ge3}\frac{\log\log n}{n\log^2 n}<\infty$ by the integral test. Also \[ b_N p_{N+1}\ll \frac{\log\log N}{N\log^3 N}\cdot (N\log N)=\frac{\log\log N}{\log^2 N}\to 0. \] Thus the partial sums $\sum_{n=3}^{N} b_n g_n$ converge, proving $5.9$. ∎

Combining Lemmas 5.4–5.5, we have the crucial approximation:

Proposition 5.6 (block approximation). There exist real numbers $e_n$ with $\sum_{n\ge3}|e_n|<\infty$ such that \[ \Sigma_n=\sum_{m=p_n}^{p_{n+1}-1}\frac{1}{m\log m} ================================================== \frac{n g_n}{p_n p_{n+1}} + e_n. \tag{5.11} \]

Step C: Convert $T(p_{x+1}-1)$ to the gap-sum

Insert $5.11$ into $5.1$: \[ T(p_{x+1}-1) ============ \sum_{n\le x}(-1)^n\frac{n g_n}{p_n p_{n+1}} + \sum_{n\le x}(-1)^n e_n. \] Since $\sum |e_n|<\infty$, the second term converges to some constant $C_2$ as $x\to\infty$. Hence \[ \sum_{n\le x}(-1)^n\frac{n g_n}{p_n p_{n+1}} ============================================ T(p_{x+1}-1)+C_3+o(1) \tag{5.12} \] for some constant $C_3$.

Step D: Replace $p_{x+1}$ by $x\log x$

Since $p_{x+1}\sim x\log x$, we have $\log\log(p_{x+1})-\log\log(x\log x)\to 0$, and thus \[ \sum_{x\log x < m \le p_{x+1}} \frac{1}{m\log m} \to 0. \] Therefore \[ T(p_{x+1}-1)=T(x\log x)+o(1). \tag{5.13} \]

Conclusion: Said–Tao equivalence

Now combine Corollary 5.3 with $5.12$–$5.13$:

Theorem 5.7 (equivalence; Said–Tao). There exists a constant $C\in\mathbb{R}$ such that \[ S(x)=\sum_{n\le x}(-1)^n\frac{n}{p_n} ===================================== \frac12,T(x\log x);+;C;+;o(1) \qquad (x\to\infty). \tag{5.14} \] In particular, \[ \sum_{n=1}^\infty (-1)^n\frac{n}{p_n} \text{ converges} \quad\Longleftrightarrow\quad \sum_{m=2}^\infty \frac{(-1)^{\pi(m)}}{m\log m} \text{ converges.} \tag{5.15} \]

This equivalence is the key obstruction noted in the literature. ([What's new][1])

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5.4 Minimal corrected sufficient condition in terms of parity discrepancy

Define the parity-discrepancy function \[ A(x):=\sum_{2\le m\le x} (-1)^{\pi(m)}. \]

By partial summation (Abel summation), for $y\ge 3$, \[ T(y)=\frac{A(y)}{y\log y}+\int_{3}^{y} A(t),d!\left(\frac{1}{t\log t}\right) + O(1), \] and \[ \left|d!\left(\frac{1}{t\log t}\right)\right| ============================================= \left(\frac{1}{t^2\log t}+\frac{1}{t^2\log^2 t}\right)dt \asymp \frac{dt}{t^2\log t}. \]

So a clean sufficient criterion is:

Proposition 5.8 (quantitative cancellation ⇒ convergence). If there exists $\varepsilon>0$ such that \[ A(x)=O!\left(\frac{x}{(\log\log x)^{1+\varepsilon}}\right), \tag{5.16} \] then (T(y)) converges as $y\to\infty$, hence the Erdős series $\sum (-1)^n n/p_n$ converges.

Proof. Under $5.16$, \[ \int_{3}^{\infty}\frac{|A(t)|}{t^2\log t},dt \ll \int_{3}^{\infty}\frac{t/(\log\log t)^{1+\varepsilon}}{t^2\log t},dt = \int_{3}^{\infty}\frac{dt}{t\log t(\log\log t)^{1+\varepsilon}} <\infty \] (because ($\int^\infty$ du/$u^{1+\varepsilon}$<$\infty$) with $u=\log\log t$). Also $A(y)/(y\log y)\to 0$. Hence (T(y)) converges. Then $5.15$ gives convergence of the Erdős series. ∎

This is “minimal corrected statement” in the sense: it isolates exactly the kind of quantitative parity cancellation that would settle the problem.

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5.5 Equidistribution alone does not imply convergence

A very tempting (but wrong) idea is: “If $\pi(m)\bmod2$ is equidistributed, then (T) should converge.” This is false in general even for abstract equidistributed sign sequences.

Proposition 5.9 (explicit equidistributed divergence example). There exists an equidistributed sequence $s(m)\in{0,1}$ $i.e. (\#\{m\le N:s(m)=0\}\sim N/2)$ such that the series \[ \sum_{m\ge 3}\frac{(-1)^{s(m)}}{m\log m} \] diverges.

Construction and proof.

1. Build a sparse set $Y\subset\mathbb{N}$ of density $0$ but with divergent $1/(m\log m)$-weight. Let \[ n_k:=2\lfloor k\log\log k\rfloor+1 \quad (k\ge 3), \] so $n_k$ is odd and strictly increasing for large (k), and $n_k\asymp k\log\log k$. Define $Y:={n_k: k\ge k_0}$ (ignore finitely many initial terms).

* Density $0$: since $n_k/k\to\infty$, the counting function satisfies $\#(Y\cap[1,N])\ll N/\log\log N = o(N)$. * Divergent weight: for large (k), \[ \frac{1}{n_k\log n_k}\asymp \frac{1}{k\log\log k\cdot \log k}, \] and $\sum_{k\ge3}\frac{1}{k\log k\log\log k}$ diverges (integral test; it grows like $\log\log\log k$). Hence \[ \sum_{m\in Y}\frac{1}{m\log m}=\infty. \tag{5.17} \]

2. Define an equidistributed (s(m)) but force divergence. Let \[ s(m):=\begin{cases} 0, & m\in Y,\ m\bmod 2, & m\notin Y. \end{cases} \] Since (Y) has density $0$, changing values on (Y) does not affect the limiting proportions; hence (s(m)) is equidistributed.

Now decompose the series: \[ \sum_{m\ge3}\frac{(-1)^{s(m)}}{m\log m} ======================================= \sum_{m\ge3}\frac{(-1)^m}{m\log m} + \sum_{m\in Y}\frac{(-1)^{s(m)}-(-1)^m}{m\log m}. \] The first sum converges by the alternating series test since $1/(m\log m)$ decreases to $0$. For $m\in Y$, (m) is odd and $s(m)=0$, so \[ (-1)^{s(m)}-(-1)^m = 1-(-1)=2, \] and therefore the second sum equals \[ 2\sum_{m\in Y}\frac{1}{m\log m}, \] which diverges by $5.17$. Thus the entire series diverges.

∎

Takeaway: even knowing “$\pi(m)\bmod 2$ is equidistributed” $i.e. (A(x)=o(x))$ would not by itself force convergence of (T), so it cannot settle the Erdős series. One needs quantitative cancellation of the type in Proposition 5.8.

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5.6 Empirical check of the equivalence (optional, not a proof)

Using a sieve up to $16$ million, I recomputed Round‑1’s $S(10^6)$ and computed $T(\lfloor 10^6\log 10^6\rfloor)$. Numerically:

* $S(10^6)= -0.01985921629844$ (matches Round‑1). * $T(\lfloor 10^6\log 10^6\rfloor)\approx -0.207173754781$.

So $S(10^6)-\tfrac12 T(\lfloor 10^6\log 10^6\rfloor)\approx 0.08373$, consistent with $5.14$ predicting $S(x)\approx \tfrac12 T(x\log x)+C$ with a slowly stabilizing constant (C).

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6) ADVERSARIAL VERIFICATION

Check 1: Did any step smuggle in “convergence of (T)”?

No. In Step D (replacing $p_{x+1}$ by $x\log x$), I bounded the difference by the absolute sum of weights: \[ \sum_{m\in (p_{x+1},,x\log x]} \frac{1}{m\log m} \le \int_{p_{x+1}}^{x\log x}\frac{dt}{t\log t} = \log\log(x\log x)-\log\log(p_{x+1})\to0, \] which requires no convergence of (T).

Check 2: The crucial error summability (Lemma 5.5)

This is the only “delicate” place. I used summation by parts in the exact identity $5.10$, plus:

* $p_{n+1}\ll n\log n$ (PNT-level bound), * $b_n-b_{n+1}\ll \log\log n/(n^2\log^3 n)$, * $\sum \log\log n/(n\log^2 n)<\infty$.

All components are standard and the final comparison series is integrable.

Check 3: Counterexample sequence (s(m))

* Equidistribution: modifications occur on (Y) of density $0$, so proportions $0/1$ remain $1/2$. * Divergence: ensured by forcing odd $m\in Y$, making the correction term exactly $2/(m\log m)$, and choosing (Y) so $\sum_{m\in Y}1/(m\log m)=\infty$.

No hidden assumptions.

Check 4: Compatibility with Round‑1 lemmas

Lemma 15.1 is consistent: the gap-based decomposition (Lemma 5.2 / Corollary 5.3) is precisely the type of “gap control” Round‑1 identified as the obstacle; we have now pinned it to parity blocks.

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7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

What is now fully resolved in this round:

* A gap-free equivalence between the Erdős series and the parity series $\sum_{m}(-1)^{\pi(m)}/(m\log m)$ (Theorem 5.7). This is a major structural reduction. * A rigorous demonstration that equidistribution of parity alone cannot prove convergence, by an explicit equidistributed divergence construction (Proposition 5.9). * A clean minimal quantitative sufficient condition (Proposition 5.8) that isolates the remaining obstruction as a bound on the discrepancy (A(x)).

What remains open:

* Proving (or disproving) a discrepancy estimate for $A(x)=\sum_{m\le x}(-1)^{\pi(m)}$ strong enough to imply convergence $e.g. (A(x)=O(x/(\log\log x)^{1+\varepsilon}))$, unconditionally.

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8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 70%

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9) REFERENCES

* Terence Tao, blog post: The convergence of an alternating series of Erdős, assuming the Hardy–Littlewood prime tuples conjecture (Aug 2023). ([What's new][1]) * MathOverflow question: A problem of Erdős on convergence of $\sum (-1)^n n/p_n$ and equidistribution of parity of $\pi(m)$ (Oct 2018). ([MathOverflow][2]) * Terence Tao, paper (published version): The convergence of an alternating series of Erdős, assuming the Hardy–Littlewood prime tuples conjecture (Comm. AMS, 2024; also arXiv:2308.07205). ([NSF Public Access Repository][3])

[1]: https://terrytao.wordpress.com/2023/08/14/the-convergence-of-an-alternating-series-of-erdos-assuming-the-hardy-littlewood-prime-tuples-conjecture/ "https://terrytao.wordpress.com/2023/08/14/the-convergence-of-an-alternating-series-of-erdos-assuming-the-hardy-littlewood-prime-tuples-conjecture/" [2]: https://mathoverflow.net/questions/313999/a-problem-of-erd%C5%91s-on-convergence-of-sum-1nn-p-n-and-equidistribution-of "https://mathoverflow.net/questions/313999/a-problem-of-erd%C5%91s-on-convergence-of-sum-1nn-p-n-and-equidistribution-of" [3]: https://par.nsf.gov/servlets/purl/10527569 "https://par.nsf.gov/servlets/purl/10527569"