../

---

Source. Terence Tao, The convergence of an alternating series of Erdős, assuming the Hardy--Littlewood prime tuples conjecture, Conjecture 1.3 and Theorem 1.4 on physical and printed p. 2, proved in Section 3 on pp. 4--12 (displays (3.1)--(3.17)), of the sixteen-page arXiv v3 PDF held by its library card, Tao (2023). The same-paper inputs are reconstructed on Lemma 3.1, Lemma 3.2 (which also holds the random sifted model and displays (3.7)--(3.8)), and relation (2.1).

Standing. This is an author-recorded conditional reconstruction. It is not an independent review, does not prove Conjecture 1.3, and does not change Problem 15's status or assign a verification tier. Every deduction of the source's Section 3 is written out; the external theorems it uses are stated below at the strength consumed and are not reproved.

Definitions and hypothesis

$p$ ranges over primes, $\mathcal P$ is the set of primes, $1_{\mathcal P}$ its indicator, $\pi(t)=\#\{p\le t\}$, and $\gamma$ is the Euler--Mascheroni constant. For a set $\mathcal H=\{h_1,\dots,h_k\}$ of distinct integers, $\nu_{\mathcal H}(p)$ and the singular series $\mathfrak S(\mathcal H)$ are as defined on the Lemma 3.2 page. Random variables are in boldface. All implied constants are absolute (they may depend on the constants $\varepsilon,C$ of the hypothesis), and "for large $x$" means for $x\ge x_0$ with $x_0$ absolute.

Hypothesis (Conjecture 1.3 of the source, p. 2). There are absolute constants $\varepsilon>0$ and $C>0$ such that for all real $x\ge10$, all integers $k\le(\log\log x)^5$, and all sets $\mathcal H=\{h_1,\dots,h_k\}\subset[0,\log^2x]$ of distinct integers,

$$\left|\sum_{n\le x}1_{\mathcal P}(n+h_1)\cdots1_{\mathcal P}(n+h_k) -\mathfrak S(\mathcal H)\int_2^x\frac{dy}{\log^ky}\right| \le Cx^{1-\varepsilon}. \tag{1.1}$$

The source takes this from Kuperberg's Conjecture 1.3 with the range of $k$ widened from $(\log\log x)^3$ to $(\log\log x)^5$, the restriction $x\ge10$ added, and the admissibility restriction dropped. Shrinking $\varepsilon$ keeps (1.1) true, since $x^{1-\varepsilon}$ grows as $\varepsilon$ shrinks; the proof assumes $\varepsilon\le1/(2e^{\gamma})$.

Imported input (Kuperberg's Theorem 1.2). For all positive integers $k$ and $h$, with no relation between them,

$$T_k(h):=\sum_{\substack{h_1,\dots,h_k\le h\\\text{distinct}}} \mathfrak S(\{h_1,\dots,h_k\}) \ll h^k\prod_{p\le k^3}\left(1-\frac1p\right)^{-k},$$

the sum over ordered $k$-tuples of distinct positive integers, with an absolute implied constant. This is the first inequality of display (5) on p. 2 of V. Kuperberg, Sums of singular series with large sets and the tail of the distribution of primes, Q. J. Math. 74 (2023), arXiv:2210.09775v2, held as Kuperberg (2023); its proof was not reread here. By Mertens' third theorem, $\prod_{p\le k^3}(1-1/p)^{-1}\ll\log k$ for $k\ge2$, so there is an absolute $C_1$ with $T_k(h)\le(C_1h\log k)^k$ for $k\ge2$; this is the form used below. Kuperberg's display continues "$\ll h^k(3\log k)^k$" and the source quotes that form; the numerical constant plays no role here.

Other imported inputs. Mertens' second and third theorems with error $O(1/\log y)$, as stated on the Lemma 3.2 page; Bertrand's postulate, $p_{n+1}\le2p_n$; the elementary bound $r!\ge(r/e)^r$; and the prime number theorem only through Mertens' theorems.

Statement

Assume Conjecture 1.3. Then the series

$$\sum_{n=2}^\infty\frac{(-1)^{\pi(n)}}{n\log n}$$

converges, and hence, by relation (2.1), so does Erdős's series $\sum_{n=1}^\infty(-1)^nn/p_n$, which answers Problem 15 affirmatively under the hypothesis.

Proof

Step 1: reduction to the partial-sum bound (3.1)

Put $a_n=(-1)^{\pi(n)}$ and $A(t)=\sum_{n\le t}a_n$. It suffices to prove

$$A(t)\ll\frac t{(\log\log t)^{1.1}}\qquad(t\text{ large}). \tag{3.1}$$

Indeed, let $f(t)=1/(t\log t)$, so that $f'(t)=-(\log t+1)/(t\log t)^2$ and $|f'(t)|\le2/(t^2\log t)$ for $t\ge e$. Summation by parts gives, for $e\le N<M$,

$$\sum_{N<n\le M}a_nf(n)=A(M)f(M)-A(N)f(N)-\int_N^MA(t)f'(t)\,dt.$$

Under (3.1), $|A(t)f(t)|\ll1/(\log t(\log\log t)^{1.1})\to0$ and

$$\int_N^M|A(t)f'(t)|\,dt \ll\int_N^M\frac{dt}{t\log t(\log\log t)^{1.1}} =\int_{\log\log N}^{\log\log M}\frac{du}{u^{1.1}},$$

which tends to $0$ as $N\to\infty$ uniformly in $M$. So the partial sums of $\sum a_nf(n)$ are Cauchy, and the series converges.

Step 2: reduction to short intervals

Let $\varepsilon$ be the constant of the hypothesis. It suffices to prove, for all large $x$,

$$\sum_{x\le n<x+x^{1-\varepsilon/2}}(-1)^{\pi(n)} \ll\frac{x^{1-\varepsilon/2}}{(\log\log x)^{1.1}}. \tag{3.1'}$$

Given (3.1'), fix a large $X$ and tile $[X^{1/2},\infty)$ by the half-open real intervals $[x_j,x_{j+1})$ with $x_0=X^{1/2}$ and $x_{j+1}=x_j+x_j^{1-\varepsilon/2}$. Let $J$ be the last index with $x_J\le X$. Then

$$A(X)=\sum_{n<X^{1/2}}a_n +\sum_{j<J}\ \sum_{x_j\le n<x_{j+1}}a_n +\sum_{x_J\le n\le X}a_n.$$

The first sum is at most $X^{1/2}$ in absolute value and the last at most $x_J^{1-\varepsilon/2}+1\le X^{1-\varepsilon/2}+1$, by the trivial bound $|a_n|\le1$. For $j<J$ we have $x_j\ge X^{1/2}$, so $\log\log x_j\ge\log\log X-\log2\ge\frac12\log\log X$ for large $X$, and (3.1') gives

$$\left|\sum_{j<J}\ \sum_{x_j\le n<x_{j+1}}a_n\right| \ll\sum_{j<J}\frac{x_{j+1}-x_j}{(\log\log x_j)^{1.1}} \ll\frac{X}{(\log\log X)^{1.1}},$$

because $\sum_{j<J}(x_{j+1}-x_j)=x_J-X^{1/2}\le X$. Since $X^{1/2}+X^{1-\varepsilon/2}+1\ll X/(\log\log X)^{1.1}$, (3.1) follows.

Step 3: probabilistic form and the van der Corput step

Fix a large $x$, let $I=\{n\in\mathbb Z:x\le n<x+x^{1-\varepsilon/2}\}$, $N_x=|I|\asymp x^{1-\varepsilon/2}$, and let $\mathbf n$ be uniform on $I$. (The source draws $\mathbf n$ from the closed interval, which differs by at most one integer and changes nothing below.) Then (3.1') reads

$$\mathbf E(-1)^{\pi(\mathbf n)}\ll\frac1{(\log\log x)^{1.1}}.$$

Shift invariance. For any function $F$ on the integers with $|F|\le1$ and any integer $h\ge0$, the sets $I$ and $I+h$ differ in at most $2h$ elements, so

$$\bigl|\mathbf EF(\mathbf n)-\mathbf EF(\mathbf n+h)\bigr|\le\frac{2h}{N_x}.$$

Introduce the length scale

$$H=\lfloor(\log\log x)^{4.4}\log x\rfloor. \tag{3.2}$$

For $0\le h\le H$ the shift error is $\ll H/x^{1-\varepsilon/2}$, which is far smaller than $(\log\log x)^{-10}$ for large $x$. Averaging over $h=1,\dots,H$,

$$\mathbf E(-1)^{\pi(\mathbf n)} =\mathbf E\,\frac1H\sum_{h=1}^H(-1)^{\pi(\mathbf n+h)} +O\!\left(\frac1{(\log\log x)^{10}}\right).$$

By the Cauchy--Schwarz inequality $|\mathbf E\mathbf Y|\le(\mathbf E|\mathbf Y|^2)^{1/2}$, it suffices to show

$$\mathbf E\left|\frac1H\sum_{h=1}^H(-1)^{\pi(\mathbf n+h)}\right|^2 \ll\frac1{(\log\log x)^{2.2}}.$$

Expanding the square and using $(-1)^{a+b}=(-1)^{a-b}$,

$$\mathbf E\left|\frac1H\sum_{h=1}^H(-1)^{\pi(\mathbf n+h)}\right|^2 =\frac1{H^2}\sum_{h,h'=1}^H \mathbf E(-1)^{\pi(\mathbf n+h)-\pi(\mathbf n+h')}.$$

For $h'\le h$, shift invariance applied to $F(m)=(-1)^{\pi(m+h-h')-\pi(m)}$ with shift $h'$ gives $\mathbf E(-1)^{\pi(\mathbf n+h)-\pi(\mathbf n+h')}=\mathbf E(-1)^{\pi(\mathbf n+(h-h'))-\pi(\mathbf n)}+O(H/N_x)$, and symmetrically for $h'>h$. Each difference $|h-h'|=\delta$ arises from at most $2H$ pairs, so the right side is at most

$$\frac2H\sum_{0\le\delta\le H} \bigl|\mathbf E(-1)^{\pi(\mathbf n+\delta)-\pi(\mathbf n)}\bigr| +O\!\left(\frac H{N_x}\right),$$

and it suffices to show

$$\sum_{0\le\delta\le H} \bigl|\mathbf E(-1)^{\pi(\mathbf n+\delta)-\pi(\mathbf n)}\bigr| \ll\frac H{(\log\log x)^{2.2}}.$$

Reduction to (3.3). For an integer $\delta$ with $1\le\delta\le H$ write $\lambda=\delta/\log x$, so that $\lambda\log x=\delta$ is an integer and $0<\lambda\le(\log\log x)^{4.4}$. (The source rounds $\lambda\log x$ to an integer; parametrizing by the integer $\delta$ makes rounding unnecessary.) The estimate to be proved is: there is an absolute $\lambda_0\ge4$ such that, for large $x$ and every integer $\delta$ with $\lambda_0\log x\le\delta\le H$,

$$\mathbf E(-1)^{\pi(\mathbf n+\delta)-\pi(\mathbf n)}\ll\frac1{\sqrt\lambda}, \qquad\lambda=\frac\delta{\log x}. \tag{3.3}$$

Given (3.3): the terms with $\delta\le\lambda_0\log x$ contribute at most $\lambda_0\log x+1\ll H/(\log\log x)^{4.4}$ by the trivial bound, and the rest contribute

$$\ll\sum_{\delta\le H}\left(\frac{\log x}\delta\right)^{1/2} \le2(H\log x)^{1/2} =2H\left(\frac{\log x}H\right)^{1/2} \ll\frac H{(\log\log x)^{2.2}},$$

since $H/\log x\asymp(\log\log x)^{4.4}$. The rest of the proof establishes (3.3). Fix such a $\delta$ and write $d=\delta$ and $\lambda=d/\log x$; note $\log x\le d\le H\le\log^2x$ for large $x$.

Step 4: Bonferroni truncation, reduction to (3.4)

Let $\mathbf N=\pi(\mathbf n+d)-\pi(\mathbf n)$, a random nonnegative integer. Choose two integers $r_0=2\lfloor(\log\log x)^{4.5}/2\rfloor$ and $r_1=r_0+1$; both are $(\log\log x)^{4.5}+O(1)$, one even and one odd. Lemma 3.1 gives

$$\sum_{k=0}^{r_1}(-2)^k\binom{\mathbf N}k\le(-1)^{\mathbf N} \le\sum_{k=0}^{r_0}(-2)^k\binom{\mathbf N}k,$$

so by linearity of expectation (3.3) follows once we show, for $r\in\{r_0,r_1\}$,

$$\sum_{k=0}^{r}(-2)^k\,\mathbf E\binom{\mathbf N}k\ll\frac1{\sqrt\lambda}. \tag{3.4}$$

Fix such an $r$. Since $\binom{\mathbf N}k$ counts the $k$-element subsets of $\{0<h\le d:\mathbf n+h\in\mathcal P\}$,

$$\mathbf E\binom{\mathbf N}k =\sum_{0<h_1<\dots<h_k\le d} \mathbf P(\mathbf n+h_1,\dots,\mathbf n+h_k\in\mathcal P).$$

Step 5: applying the hypothesis, reduction to (3.5)

Let $0\le k\le r$ and $\mathcal H=\{h_1,\dots,h_k\}$ with $0<h_1<\dots<h_k\le d$. Let $y_1=\lceil x\rceil-1$ and $y_2=\max I$, so that $I=\{y_1<n\le y_2\}$, $y_2-y_1=N_x$, and $y_1\ge x-1\ge10$. For large $x$, $k\le(\log\log x)^{4.5}+O(1)\le(\log\log y_1)^5$ and $d\le H\le\log^2y_1$, so (1.1) applies at $y_1$ and at $y_2$, and subtracting,

$$\sum_{n\in I}\prod_{i=1}^k1_{\mathcal P}(n+h_i) =\mathfrak S(\mathcal H)\int_{y_1}^{y_2}\frac{dt}{\log^kt} +O(x^{1-\varepsilon}).$$

For $t\in[y_1,y_2]$, $\log t=\log x+O(x^{-\varepsilon/2})$, so $\log^{-k}t=\log^{-k}x\,(1+O(kx^{-\varepsilon/2}))=\log^{-k}x\,(1+O(x^{-\varepsilon/3}))$, and the integral is $N_x\log^{-k}x\,(1+O(x^{-\varepsilon/3}))$. Dividing by $N_x$,

$$\mathbf P(\mathbf n+h_1,\dots,\mathbf n+h_k\in\mathcal P) =\frac{\mathfrak S(\mathcal H)}{\log^kx}\left(1+O(x^{-\varepsilon/3})\right) +O(x^{-\varepsilon/2}).$$

The left side is at most $1$, so $\mathfrak S(\mathcal H)/\log^kx\le3$ for large $x$ (a fill: the source says "routine manipulations"), and therefore

$$\mathbf P(\mathbf n+h_1,\dots,\mathbf n+h_k\in\mathcal P) =\frac{\mathfrak S(\mathcal H)}{\log^kx}+O(x^{-\varepsilon/3}).$$

The source writes the error as $O(x^{-\varepsilon/2})$; only a power saving is used. Summing over the at most $\binom dk\le d^k$ sets $\mathcal H$ and over $k\le r$, the error contributes at most $(r+1)(2d)^rx^{-\varepsilon/3}$, and $(2d)^r=\exp(r\log(2d))\le\exp(O((\log\log x)^{5.5}))=x^{o(1)}$, so this is $O(x^{-\varepsilon/4})$, negligible against $1/\sqrt\lambda$. Hence (3.4) follows from

$$\sum_{k=0}^{r}\frac{(-2)^k}{\log^kx} \sum_{0<h_1<\dots<h_k\le d}\mathfrak S(\mathcal H)\ll\frac1{\sqrt\lambda}. \tag{3.5}$$

The source remarks that Kuperberg's mean-value estimates would handle each $k$ separately only for $k$ below about $(\log\log x)^{0.5}$, so the oscillation in $k$ must be kept; the next step does so by passing to the random sifted model.

Step 6: the sieve cutoff and passage to the model

Let $z$ be the smallest prime with

$$\prod_{p\le z}\left(1-\frac1p\right)\le\frac1{\log x}.$$

The source says "largest prime"; since the product decreases in $z$, the condition holds for every sufficiently large prime, and the reading "smallest" is the one under which the source's display (3.6) holds. If $z'$ is the prime preceding $z$ then $\prod_{p\le z'}(1-1/p)>1/\log x$, so $\prod_{p\le z}(1-1/p)>(1-1/z)/\log x$, and hence

$$\prod_{p\le z}\left(1-\frac1p\right)=\frac1{\log x}+O\!\left(\frac1{z\log x}\right).$$

Comparing with Mertens' third theorem, $e^{-\gamma}/\log z\,(1+O(1/\log z))=(1/\log x)(1+O(1/z))$, so $\log z=e^{-\gamma}\log x\,(1+O(1/\log x))$, that is $z\asymp x^{1/e^\gamma}$, and

$$\prod_{p\le z}\left(1-\frac1p\right)=\frac1{\log x}+O(x^{-1/e^\gamma}). \tag{3.6}$$

In particular $z>d$ for large $x$. Run the random sifted model of the Lemma 3.2 page with this $d$ and $z$: sets $\boldsymbol{\mathcal S}_w\subset(0,d]$ and counts $\mathbf S_w$ for $w\le z$. For $k\le r$ and $\mathcal H$ as in Step 5, display (3.8) at level $w=z$, whose hypothesis $k^2\le z$ holds for large $x$ because $k\le(\log\log x)^{4.5}+O(1)$ and $z\asymp x^{1/e^\gamma}$, gives

$$\mathbf P(h_1,\dots,h_k\in\boldsymbol{\mathcal S}_z) =\mathfrak S(\mathcal H) \left(\prod_{p\le z}\left(1-\frac1p\right)\right)^k \left(1+O\!\left(\frac{k^2}z\right)\right).$$

By (3.6), $\left(\prod_{p\le z}(1-1/p)\right)^k=\log^{-k}x\,(1+O(x^{-1/e^\gamma}\log x))^k=\log^{-k}x\,(1+O(x^{-1/(2e^\gamma)}))$, and $k^2/z\ll x^{-1/(2e^\gamma)}$. Using $\mathfrak S(\mathcal H)/\log^kx\le3$ from Step 5 and $\varepsilon\le1/(2e^\gamma)$,

$$\mathbf P(h_1,\dots,h_k\in\boldsymbol{\mathcal S}_z) =\frac{\mathfrak S(\mathcal H)}{\log^kx}+O(x^{-\varepsilon}). \tag{3.9}$$

Summing as in Step 5, the error contributes $O(x^{-\varepsilon/2})$ to the left side of (3.5), so (3.5) follows from

$$\sum_{k=0}^{r}(-2)^k\sum_{0<h_1<\dots<h_k\le d} \mathbf P(h_1,\dots,h_k\in\boldsymbol{\mathcal S}_z) =\sum_{k=0}^{r}(-2)^k\,\mathbf E\binom{\mathbf S_z}k \ll\frac1{\sqrt\lambda},$$

the equality because $\binom{\mathbf S_z}k$ counts the $k$-subsets of $\boldsymbol{\mathcal S}_z$. By the two-sided form of Lemma 3.1 applied to $N=\mathbf S_z$ (here $r\ge1$),

$$\left|\sum_{k=0}^{r}(-2)^k\binom{\mathbf S_z}k-(-1)^{\mathbf S_z}\right| \le2^r\binom{\mathbf S_z}r,$$

so it suffices to establish

$$\mathbf E(-1)^{\mathbf S_z}\ll\frac1{\sqrt\lambda} \tag{3.10}$$

and

$$2^r\,\mathbf E\binom{\mathbf S_z}r\ll\frac1{\sqrt\lambda}. \tag{3.11}$$

Step 7: the tail term (3.11)

By (3.9) with $k=r$ and the same error accounting,

$$2^r\,\mathbf E\binom{\mathbf S_z}r =\frac{2^r}{\log^rx}\sum_{0<h_1<\dots<h_r\le d}\mathfrak S(\mathcal H) +O(x^{-\varepsilon/2}).$$

The sum over increasing $r$-tuples is $T_r(d)/r!$ in Kuperberg's notation, so the imported Theorem 1.2 and $r!\ge(r/e)^r$ give

$$\frac{2^r}{\log^rx}\cdot\frac{T_r(d)}{r!} \le\frac{(2C_1d\log r)^r}{r!\,\log^rx} =\frac{(2C_1\lambda\log r)^r}{r!} \le\left(\frac{2eC_1\lambda\log r}r\right)^r.$$

Since $\lambda\le(\log\log x)^{4.4}$, $r\ge(\log\log x)^{4.5}-2$ and $\log r\le5\log\log\log x$, the base is $\ll(\log\log\log x)(\log\log x)^{-0.1}\to0$, so for large $x$ it is at most $1/2$, and the quantity is at most $2^{-r}\le2^{2-(\log\log x)^{4.5}}$, which is $\ll(\log\log x)^{-2.2}\le1/\sqrt\lambda$. This proves (3.11).

Step 8: the bias recursion and (3.10)

One sifting step. Let $q^-<q$ be consecutive primes with $d<q^-$ and $q\le z$. The set $\boldsymbol{\mathcal S}_q$ is obtained from $\boldsymbol{\mathcal S}_{q^-}$ by removing the elements congruent to $\mathbf a_q$ modulo $q$; $\mathbf a_q$ is uniform modulo $q$ and independent of $(\mathbf a_p)_{p\le q^-}$, hence of $\boldsymbol{\mathcal S}_{q^-}$. Since $\boldsymbol{\mathcal S}_{q^-}\subset(0,d]$ and $q>d$, its elements lie in distinct residue classes modulo $q$, so conditionally on $\boldsymbol{\mathcal S}_{q^-}$ exactly one element is removed with probability $\mathbf S_{q^-}/q$ and none otherwise. Therefore

$$\mathbf E\bigl[(-1)^{\mathbf S_q}\,\big|\,\boldsymbol{\mathcal S}_{q^-}\bigr] =\left(1-\frac{\mathbf S_{q^-}}q\right)(-1)^{\mathbf S_{q^-}} +\frac{\mathbf S_{q^-}}q(-1)^{\mathbf S_{q^-}-1} =\left(1-\frac{2\mathbf S_{q^-}}q\right)(-1)^{\mathbf S_{q^-}},$$

and by the law of total expectation

$$\mathbf E(-1)^{\mathbf S_q} =\mathbf E\left(1-\frac{2\mathbf S_{q^-}}q\right)(-1)^{\mathbf S_{q^-}}.$$

Write $\mu=\mathbf E\,\mathbf S_{q^-}$ and split $\mathbf S_{q^-}=\mu+(\mathbf S_{q^-}-\mu)$:

$$\mathbf E(-1)^{\mathbf S_q} =\left(1-\frac{2\mu}q\right)\mathbf E(-1)^{\mathbf S_{q^-}} -\frac2q\,\mathbf E\bigl[(\mathbf S_{q^-}-\mu)(-1)^{\mathbf S_{q^-}}\bigr].$$

The factor $1-2\mu/q$ is positive: by (3.12), $\mu=d\prod_{p\le q^-}(1-1/p)\le d/3\le q^-/3<q/3$, since $q^-\ge3$. So by the triangle inequality

$$\bigl|\mathbf E(-1)^{\mathbf S_q}\bigr| \le\left(1-\frac{2\mu}q\right)\bigl|\mathbf E(-1)^{\mathbf S_{q^-}}\bigr| +\frac2q\,\mathbf E|\mathbf S_{q^-}-\mu|.$$

This is the source's key observation: the factor is slightly less than one, so each sifting step damps the bias. By the Cauchy--Schwarz inequality and (3.13) at $w=q^-$ (valid as $d\le q^-\le z$ and $d\ge\log x$ exceeds the constant $d_0$ of Lemma 3.2 for large $x$), $\mathbf E|\mathbf S_{q^-}-\mu|\le\mathbf{Var}(\mathbf S_{q^-})^{1/2}\ll(d/\log q^-)^{1/2}\ll(d/\log q)^{1/2}$, using Bertrand's postulate $q\le2q^-$, so $\log q^-\ge\log q-\log2\ge\frac12\log q$. By (3.12) and $\log q^-=\log q+O(1)$,

$$\mu=\frac{d}{e^\gamma\log q^-}\left(1+O\!\left(\frac1{\log q^-}\right)\right) =\frac{d}{e^\gamma\log q}\left(1+O\!\left(\frac1{\log q}\right)\right).$$

Bounding $1-2\mu/q\le\exp(-2\mu/q)$ gives the recursive inequality

$$\bigl|\mathbf E(-1)^{\mathbf S_q}\bigr| \le\rho(q)\bigl|\mathbf E(-1)^{\mathbf S_{q^-}}\bigr|+e(q),$$ $$\rho(q)=\exp\left(-\frac{2d}{e^\gamma q\log q} +O\!\left(\frac d{q\log^2q}\right)\right), \qquad e(q)\ll\frac1q\left(\frac d{\log q}\right)^{1/2}.$$

The source indexes the exponent and the error by the smaller prime $q^-$ (its $p_n$) instead of $q$ (its $p_{n+1}$). The error term and the $O$-term agree with the forms above up to constants by the Bertrand step; the main term does not, since $1/(q^-\log q^-)-1/(q\log q)$ is of order $(q-q^-)/(q^2\log q)$, and the source's form needs the prime-gap bound $q-q^-\ll q/\log q$, a consequence of the prime number theorem and not of Bertrand's postulate. The recursion above avoids this by keeping $q$; in the product $\alpha_w$ the two indexings differ by a bounded factor, since the differences of the decreasing function $1/(t\log t)$ over consecutive primes telescope to at most $1/(q_0\log q_0)$.

Iteration. Let $q_0<q_1<\dots<q_M=z$ be the primes in $(d,z]$, and $b_j=|\mathbf E(-1)^{\mathbf S_{q_j}}|$. Induction on $M$ from $b_j\le\rho(q_j)b_{j-1}+e(q_j)$ (the source's footnote 8: a discrete Gronwall inequality) gives

$$b_M\le b_0\prod_{j=1}^M\rho(q_j)+\sum_{j=1}^Me(q_j)\prod_{i=j+1}^M\rho(q_i).$$

With the trivial bound $b_0\le1$ and, for $d\le w\le z$,

$$\alpha_w:=\prod_{w<q\le z}\rho(q) =\exp\left(-\sum_{w<q\le z}\left(\frac{2d}{e^\gamma q\log q} +O\!\left(\frac d{q\log^2q}\right)\right)\right)$$

(an empty product being $1$), this reads

$$\mathbf E(-1)^{\mathbf S_z}\ll\alpha_{q_0}+\sum_{d<q\le z}\frac{\alpha_q}q \left(\frac d{\log q}\right)^{1/2}.$$

Moreover $\alpha_{q_0}=\alpha_d/\rho(q_0)\ll\alpha_d$, because $\rho(q_0)\ge\exp(-2e^{-\gamma}/\log q_0-O(1/\log^2q_0))\gg1$ as $d<q_0$.

Evaluating $\alpha_w$. Let $R(y)=\sum_{p\le y}1/p=\log\log y+B+E(y)$ with $E(y)=O(1/\log y)$ (Mertens' second theorem). For $d\le w\le z$, Stieltjes integration against $R$ with the decreasing function $1/\log t$ gives

$$\sum_{w<q\le z}\frac1{q\log q} =\int_w^z\frac{dt}{t\log^2t} +\left[\frac{E(t)}{\log t}\right]_w^z +\int_w^z\frac{E(t)}{t\log^2t}\,dt =\frac1{\log w}-\frac1{\log z}+O\!\left(\frac1{\log^2w}\right),$$

and likewise $\sum_{w<q\le z}1/(q\log^2q)\ll1/\log^2w$. Hence

$$\alpha_w=\exp\left(-\frac{2d}{e^\gamma\log w}+\frac{2d}{e^\gamma\log z} +O\!\left(\frac d{\log^2w}\right)\right) \qquad(d\le w\le z). \tag{3.15}$$

Note $2d/(e^\gamma\log z)=2\lambda(1+O(1/\log x))\le3\lambda$ by the estimate for $\log z$, and the $O$-term is $O(1/\log w)$ times the first term, hence at most half of it for large $x$, since $\log w\ge\log d\ge\log\log x$.

The starting weight. At $w=d$, $\log d\le2\log\log x$, so the exponent in (3.15) is at most $-\frac{d}{e^\gamma\log d}+3\lambda\le-\frac{d}{2e^\gamma\log\log x}+3\lambda\le-\frac{\log x}{4e^\gamma\log\log x}$ for large $x$ (as $d\ge\log x$ and $\lambda\le d/\log x$). Thus $\alpha_d\le\exp(-c\log x/\log\log x)$ with $c>0$ absolute, which is $\ll(\log\log x)^{-2.2}\le1/\sqrt\lambda$. It remains to show

$$\sum_{d<q\le z}\frac{\alpha_q}q\left(\frac d{\log q}\right)^{1/2} \ll\frac1{\sqrt\lambda}.$$

Small primes $d<q\le x^{1/(100\log\log x)}$. Here $\log q\le\log x/(100\log\log x)$, so $\frac{2d}{e^\gamma\log q}\ge200e^{-\gamma}\lambda\log\log x$, and the exponent in (3.15) is at most $-100e^{-\gamma}\lambda\log\log x+3\lambda\le-50\lambda\log\log x\le-50\log\log x$, using $100e^{-\gamma}>56$ and $\lambda\ge\lambda_0\ge1$. So $\alpha_q\le\log^{-50}x$, while crudely $(d/\log q)^{1/2}\le d^{1/2}\le\log x$. By Mertens' second theorem the contribution of this range is

$$\ll\log^{-49}x\sum_{q\le x}\frac1q\ll\log^{-48}x\ll\frac1{\sqrt\lambda}.$$

(The source's bounds are $\log^{-10}x$ and $\log^{-8}x$; the exponents are immaterial.) It remains to show

$$\sum_{x^{1/(100\log\log x)}\le q\le z}\frac{\alpha_q}q \left(\frac d{\log q}\right)^{1/2}\ll\frac1{\sqrt\lambda}. \tag{3.16}$$

Large primes. For $q$ in the range of (3.16), $\log q\ge\log x/(100\log\log x)$, so $d/\log^2q\le10^4\lambda(\log\log x)^2/\log x\le1$ for large $x$; the $O$-term in (3.15) is bounded, and

$$\alpha_q\ll\exp\left(-\frac2{e^\gamma} \left(\frac d{\log q}-\frac d{\log z}\right)\right).$$

Put $\theta=d/\log z=e^\gamma\lambda(1+O(1/\log x))$, so $\lambda\le\theta\le2\lambda$ for large $x$. For each $q$ in the range let $m=\lfloor d/\log q\rfloor$, so that

$$m\le\frac d{\log q}<m+1, \tag{3.17}$$

and $\theta-1\le m\le100\lambda\log\log x\le100(\log\log x)^{5.4}$; in particular $m\ge1$ as $\theta\ge\lambda_0\ge4$. Then $\alpha_q\ll\exp(-\frac2{e^\gamma}(m-\theta))$ and $(d/\log q)^{1/2}\le(m+1)^{1/2}\le(2m)^{1/2}$. The primes with a given $m$ satisfy $d/(m+1)<\log q\le d/m$, that is $e^{d/(m+1)}<q\le e^{d/m}$, and Mertens' second theorem gives

$$\sum_{q:\,\lfloor d/\log q\rfloor=m}\frac1q \le\log\frac{d/m}{d/(m+1)}+O\!\left(\frac{m+1}d\right) =\log\left(1+\frac1m\right)+O\!\left(\frac{m+1}d\right)\ll\frac1m,$$

because $(m+1)^2\ll\lambda^2(\log\log x)^2\le(\log\log x)^{10.8}$, which is at most $\log x\le d$ for large $x$, so $(m+1)/d\ll1/(m+1)$. (This is the source's remark that (3.17) confines $\log\log q$ to an interval of length $O(1/m)$ that still contains a dyadic range.) Hence the left side of (3.16) is

$$\ll\sum_{m\ge\theta-1}\frac1{m^{1/2}} \exp\left(-\frac2{e^\gamma}(m-\theta)\right).$$

Writing $m=m_0+j$ with $m_0=\lceil\theta-1\rceil$ and $j\ge0$, we have $m-\theta\ge j-1$ and $m\ge m_0\ge\theta-1\ge\theta/2$, so the sum is

$$\ll\theta^{-1/2}\sum_{j\ge0}e^{-2e^{-\gamma}j}\ll\theta^{-1/2} \asymp\frac1{\sqrt\lambda}.$$

This proves (3.16), hence (3.10).

Step 9: conclusion

Steps 7 and 8 give (3.10) and (3.11); Step 6 turns them into (3.5); Step 5 turns (3.5) into (3.4) for both parities of $r$; Step 4 turns that into (3.3); Step 3 turns (3.3), together with the trivial bound for $\delta\le\lambda_0\log x$, into (3.1'); Step 2 gives (3.1); and Step 1 gives the convergence of $\sum_{n\ge2}(-1)^{\pi(n)}/(n\log n)$. Relation (2.1) then gives the convergence of $\sum_{n\ge1}(-1)^nn/p_n$. All thresholds ("large $x$") and implied constants are absolute, given the constants of Conjecture 1.3. This proves Theorem 1.4.

Compilation notes

• The sieve cutoff is read as the smallest prime with $\prod_{p\le z}(1-1/p)\le1/\log x$; the source says "largest", which would make the condition vacuous. Only (3.6) and $z\asymp x^{1/e^\gamma}$ are used.
• The bound $\mathfrak S(\mathcal H)/\log^kx\le3$ in Step 5, derived from (1.1) itself, replaces the source's "routine manipulations"; the error exponents $\varepsilon/3$, $\varepsilon/4$ and the exponents $50$, $48$ of $\log x$ in Step 8 differ from the source's, which only need to be power or logarithmic savings.
• The recursion is indexed by the larger of the two consecutive primes; the source indexes it by the smaller one, which needs a prime-gap bound beyond Bertrand's postulate; the derivation here does not.
• The source's Remark 3.3 (a convergence rate $O((\log\log x)^{-0.1})$ for both partial sums under the hypothesis), Section 4 (the extension to $\sum z^nn/p_n$ for unimodular $z\ne1$, with a sketched proof), and Section 5 (further series) are not reconstructed.

Boundary. The only conditional input is Conjecture 1.3, used in Step 5 at $x\ge10$, $k\le(\log\log x)^{4.5}+O(1)$ and shifts in $(0,\log^2x]$. The imported unconditional inputs are Kuperberg's Theorem 1.2 (Step 7), the pair singular-series average (through Lemma 3.2), Mertens' theorems, Bertrand's postulate and the prime number theorem (through relation (2.1)). The hypothesis remains unproved, and this page establishes only the implication.