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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 Oct 28 '18 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 Oct 28 '18 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 Oct 29 '18 at 9:43
  • $\begingroup$ @SylvainJULIEN $|E_n| - |O_n|$ is certainly not bounded, because there are arbitrarily large gaps in the primes. $\endgroup$ – Sean Eberhard Oct 29 '18 at 11:55

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