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Source. Terence Tao, The convergence of an alternating series of Erdős, assuming the Hardy--Littlewood prime tuples conjecture, Section 2 (displays (2.1)--(2.3)), physical and printed pp. 3--4 of the sixteen-page arXiv v3 PDF held by its library card, Tao (2023). The source credits the equivalence to an unpublished observation of Said (its footnote 3 points to MathOverflow question 313999) and supplies the proof "for the convenience of the reader".

Standing. This is an author-recorded reconstruction of a source argument. It is not an independent review, changes no status and assigns no tier. The result is unconditional; its only external input is the prime number theorem in the form stated below. The source states displays (2.1)--(2.3), the intermediate displays reproduced in Steps 1--4, 6 and 7, and the summation-by-parts bound that Step 8 makes explicit; the tail estimate in Step 4, the explicit constant in Step 5, the calculation in Step 7, the derivative bound and the integral comparison in Step 8, the derivation of the prime number theorem form from $\pi(t)$, and Step 9 are supplied by this reconstruction where the source writes 'from the prime number theorem and subdivision of the $m$ variable', 'after some calculation', 'from summation by parts and the prime number theorem' and 'clearly follows'.

Definitions

$p_n$ is the $n$th prime and $\pi(t)=\#\{p\le t\}$. For real $x\ge1$ and $y\ge2$ put

$$A(x)=\sum_{n\le x}\frac{(-1)^nn}{p_n}, \qquad B(y)=\sum_{2\le m\le y}\frac{(-1)^{\pi(m)}}{m\log m}.$$

Question 1.1 of the source asks whether $A(x)$ converges as $x\to\infty$ (this is Problem 15); Question 1.2 asks whether $B(y)$ converges as $y\to\infty$.

Imported input (prime number theorem). For $n\ge10$,

$$p_n=n\log n\left(1+O\!\left(\frac{\log\log n}{\log n}\right)\right),$$

with an absolute implied constant. This follows from $\pi(t)=\frac t{\log t}(1+O(1/\log t))$: it gives $p_n=n\log p_n\,(1+O(1/\log n))$, and $\log p_n=\log n+O(\log\log n)$. Also $p_n/n\to\infty$ and $n/p_n\to0$.

Statement

There is an absolute constant $C$ such that

$$A(x)=\tfrac12B(x\log x)+C+o(1)\qquad(x\to\infty). \tag{2.1}$$

Consequently $A(x)$ converges as $x\to\infty$ if and only if $B(y)$ converges as $y\to\infty$.

Proof

Step 1: averaging with the shifted sum

Reindexing $m=n+1$,

$$\sum_{n\le x}\frac{(-1)^{n+1}(n+1)}{p_{n+1}} =\sum_{2\le m\le\lfloor x\rfloor+1}\frac{(-1)^mm}{p_m} =A(x)-\frac{(-1)^1\cdot1}{p_1}+\frac{(-1)^{\lfloor x\rfloor+1}(\lfloor x\rfloor+1)}{p_{\lfloor x\rfloor+1}} =A(x)+\frac12+o(1),$$

since $p_1=2$ and $n/p_n\to0$. Thus $A(x)=-\frac12+\sum_{n\le x}\frac{(-1)^{n+1}(n+1)}{p_{n+1}}+o(1)$. Averaging this with the identity $A(x)=A(x)$,

$$A(x)=-\frac14+\frac12\sum_{n\le x}\left(\frac{(-1)^nn}{p_n} +\frac{(-1)^{n+1}(n+1)}{p_{n+1}}\right)+o(1).$$

Step 2: the summand identity (2.2)

For each $n$,

$$\frac{(-1)^nn}{p_n}+\frac{(-1)^{n+1}(n+1)}{p_{n+1}} =(-1)^n\frac{np_{n+1}-(n+1)p_n}{p_np_{n+1}} =(-1)^n\frac{n(p_{n+1}-p_n)-p_n}{p_np_{n+1}} =\frac{(-1)^nn(p_{n+1}-p_n)}{p_np_{n+1}}-\frac{(-1)^n}{p_{n+1}}.$$

Step 3: the alternating tail

Since $1/p_{n+1}$ decreases to $0$, the alternating series test gives an absolute constant $C_0$ with $\sum_{n\le x}\frac{(-1)^n}{p_{n+1}}=C_0+o(1)$. Hence

$$A(x)=-\frac14-\frac{C_0}2 +\frac12\sum_{n\le x}\frac{(-1)^nn(p_{n+1}-p_n)}{p_np_{n+1}}+o(1).$$

Step 4: regrouping $B$ by prime gaps

The intervals $[p_n,p_{n+1})$ for $n\le x$ tile $[2,p_{\lfloor x\rfloor+1})$, so

$$\sum_{n\le x}\ \sum_{p_n\le m<p_{n+1}}\frac{(-1)^{\pi(m)}}{m\log m} =B\bigl(p_{\lfloor x\rfloor+1}-1\bigr).$$

By the imported prime number theorem, $p_{\lfloor x\rfloor+1}=x\log x\,(1+o(1))$. If $a<b$ are the smaller and larger of $x\log x$ and $p_{\lfloor x\rfloor+1}-1$, then $b/a\to1$ and

$$|B(b)-B(a)|\le\sum_{a<m\le b}\frac1{m\log m} \le\frac1{a\log a}+\int_a^b\frac{dt}{t\log t} =\frac1{a\log a}+\log\frac{\log b}{\log a} =o(1),$$

because $\log b-\log a=\log(b/a)=o(1)$. Therefore

$$B(x\log x)=\sum_{n\le x}\ \sum_{p_n\le m<p_{n+1}} \frac{(-1)^{\pi(m)}}{m\log m}+o(1).$$

Step 5: reduction to an absolutely convergent series

Suppose the series

$$\sum_{n=1}^\infty\left(\sum_{p_n\le m<p_{n+1}}\frac{(-1)^{\pi(m)}}{m\log m} -\frac{(-1)^nn(p_{n+1}-p_n)}{p_np_{n+1}}\right)$$

converges absolutely, to a sum $D$. Then its partial sums up to $x$ are $D+o(1)$, and Steps 3 and 4 give

$$A(x)=-\frac14-\frac{C_0}2-\frac D2+\frac12B(x\log x)+o(1),$$

which is (2.1) with $C=-\frac14-\frac{C_0}2-\frac D2$. Since $\pi(m)=n$ for $p_n\le m<p_{n+1}$, the $n$th term of the series equals $(-1)^n$ times a real number, and absolute convergence is equivalent to

$$\sum_{n=10}^\infty\left|\sum_{p_n\le m<p_{n+1}}\frac1{m\log m} -\frac{n(p_{n+1}-p_n)}{p_np_{n+1}}\right|<\infty. \tag{2.3}$$

The starting index $10$ is arbitrary (the source's footnote 4); the first nine terms are finite.

Step 6: the mean value point

The function $g(t)=1/(t\log t)$ is continuous and decreasing on $[2,\infty)$. The average of $g$ over the $p_{n+1}-p_n$ integers $m\in[p_n,p_{n+1})$ lies between $g(p_{n+1}-1)$ and $g(p_n)$, so by the intermediate value theorem there is $x_n\in[p_n,p_{n+1}-1]$ with

$$\sum_{p_n\le m<p_{n+1}}\frac1{m\log m}=\frac{p_{n+1}-p_n}{x_n\log x_n}.$$

Step 7: comparing $1/(x_n\log x_n)$ with $n/(p_np_{n+1})$

Let $n\ge10$ and $\eta_n=\log\log n/\log n$. The imported prime number theorem gives $p_n=n\log n\,(1+O(\eta_n))$, and also $p_{n+1}=(n+1)\log(n+1)\,(1+O(\eta_{n+1}))=n\log n\,(1+O(\eta_n))$, since $(n+1)\log(n+1)=n\log n\,(1+O(1/n))$ and $\eta_{n+1}\asymp\eta_n$. As $p_n\le x_n<p_{n+1}$, also $x_n=n\log n\,(1+O(\eta_n))$, and then $\log x_n=\log n+\log\log n+O(\eta_n)=\log n\,(1+O(\eta_n))$. Hence

$$\frac1{x_n\log x_n}=\frac{1+O(\eta_n)}{n\log^2n}, \qquad \frac n{p_np_{n+1}}=\frac{1+O(\eta_n)}{n\log^2n},$$

and so

$$\frac1{x_n\log x_n}=\frac n{p_np_{n+1}} +O\!\left(\frac{\log\log n}{n\log^3n}\right)\qquad(n\ge10).$$

Step 8: summing against the prime gaps

By Steps 6 and 7 the $n$th term of (2.3) is $\ll w_n(p_{n+1}-p_n)$ with $w_n=\log\log n/(n\log^3n)$, so it suffices to show $\sum_{n\ge10}w_n(p_{n+1}-p_n)<\infty$. Summation by parts gives, for $N\ge11$,

$$\sum_{n=10}^{N}w_n(p_{n+1}-p_n) =w_Np_{N+1}-w_{10}p_{10}-\sum_{n=11}^{N}(w_n-w_{n-1})p_n.$$

The function $w(t)=\log\log t/(t\log^3t)$ has

$$w'(t)=\frac1{t^2\log^4t}-\frac{\log\log t\,(\log t+3)}{t^2\log^4t}, \qquad |w'(t)|\ll\frac{\log\log t}{t^2\log^3t}\quad(t\ge10),$$

so $|w_n-w_{n-1}|\ll\log\log n/(n^2\log^3n)$, and with $p_n\ll n\log n$,

$$\sum_{n\ge11}|w_n-w_{n-1}|\,p_n \ll\sum_{n\ge11}\frac{\log\log n}{n\log^2n}<\infty,$$

the last series converging by comparison with $\int\frac{\log\log t}{t\log^2t}\,dt=\int ue^{-u}\,du$ under $u=\log\log t$. Also $w_Np_{N+1}\ll\log\log N/\log^2N\to0$. So the partial sums converge; since the terms $w_n(p_{n+1}-p_n)$ are nonnegative, the series converges. This proves (2.3), hence the absolute convergence in Step 5, hence (2.1).

Step 9: the equivalence

If $B(y)$ converges as $y\to\infty$, then (2.1) shows $A(x)$ converges. Conversely, suppose $A(x)$ converges as $x\to\infty$ through the integers. By (2.1), $B(x\log x)$ converges along the integers $x$. For real $y\to\infty$ choose the integer $x$ with $x\log x\le y<(x+1)\log(x+1)$; the number of integers $m$ in $(x\log x,y]$ is at most $(x+1)\log(x+1)-x\log x+1\ll\log x$, each contributing at most $1/(x\log x\cdot\log(x\log x))$ to $B$, so $|B(y)-B(x\log x)|\ll1/(x\log x)\to0$, and $B(y)$ converges. This proves the statement.

Boundary. Only the prime number theorem enters, in the form stated above. The source's Remark 2.1, that the $o(1)$ in (2.1) can be taken to be $O(\log\log x/\log x)$, is stated without proof in the source and is not reconstructed here. The Theorem 1.4 reconstruction uses this page only through the equivalence, to pass from Question 1.2 to Question 1.1.