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Erdős asked1 whether the series

$$\sum_{n=1}^\infty \frac{(-1)^n n}{p_n}$$ converges.

Here, $p_n$ denotes the n-th prime. I can show that this series converges simultaneously with the series $\sum_{m=2}^\infty \frac{(-1)^{\pi(m)}}{m \log m} $ by using the prime number theorem and estimating the difference between $ \frac{n}{p_n} $ and $ \frac{n+1}{p_{n+1}}$ for odd and even $n$. Hence, the problem comes down to understanding the equidistribution of the parity of the prime counting function $\pi(m)$.

Let $E_n = \{ m \leq n : \pi(m) \equiv 0 \mod 2 \}$ and $O_n = \{ m \leq n : \pi(m) \equiv 1 \mod 2 \}$. Then one naturally asks:

Is $\lim_{n \to \infty} \frac{|E_n|}{n} = \lim_{n \to \infty}\frac{|O_n|}{n}=\frac{1}{2}$?

If this result is true, can we prove convergence?

1See, for example: Guy R.K. Unsolved problems in number theory (2nd ed., Springer, 1994), page 203, E7 or Steven R.Finch: Mathematical Constants (Cambridge University Press, 2003), page 96.

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  • $\begingroup$ @GerhardPaseman, Mustafa: I suggest to erase all previous comments, obsolete now $\endgroup$
    – YCor
    Commented Oct 28, 2018 at 16:20
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    $\begingroup$ The convergence cannot be formally deduced if the $1/2$ distribution result holds. To be more precise, it is an exercise that there exists a $\{0,1\}$-equidistributed (in the given sense) sequence $(s(n))$ such that $\sum \frac{(-1)^{s(n)}}{n\log n}$ does not converge. $\endgroup$
    – YCor
    Commented Oct 28, 2018 at 16:24
  • $\begingroup$ Intuitively, the convergence requires the sequence $ (-1)^{\pi(n)} $ to be "close to" $ (-1)^{n} $. So perhaps one should find a tight upper bound (probably an $ O(1) $ ) for the quantity $ \vert\vert E(n)\vert-\vert O(n)\vert\vert $. $\endgroup$
    – Sylvain JULIEN
    Commented Oct 29, 2018 at 9:43
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    $\begingroup$ @User Start from $s_0(n)=n$ (modulo 2). This one is equidistributed and the series converges. Now consider $X=\{2\lfloor n\log n\rfloor:n\ge 4\}$ and $Y=X\cup (X+1)$. Then $Y$ has density zero but $(1/n\log n)$ is already non-summable on $Y$, while the restriction of $(-1)^n/(n\log n)$ converges for partials sums restricted to $Y$. So, define $s(n)=s_0(n)$ for $n\notin Y$ and $s(n)=0$ for $n\in Y$. Since $s=s_0$ on a set of density 1, the equidistribution still holds. But the contribution on $Y$ (and convergence outside $Y$) will force divergence of the series $(-1)^{s(n)}/n\log n$. $\endgroup$
    – YCor
    Commented Mar 3, 2022 at 10:34
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    $\begingroup$ I was able to use the equivalence mentioned by the OP in a recent paper on this topic. arxiv.org/abs/2308.07205 $\endgroup$
    – Terry Tao
    Commented Sep 5, 2023 at 17:10

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