Erdős asked<sup>1</sup> 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 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?

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<sup>1</sup>See, for example: 
Guy R.K. *Unsolved problems in number theory* (2nd ed., Springer, 1994), [page 203, E7](https://books.google.com/books?id=EbLzBwAAQBAJ&pg=PA203)
or Steven R.Finch: *Mathematical Constants* (Cambridge University Press, 2003), [page 96](https://books.google.com/books?id=Pl5I2ZSI6uAC&pg=PA96).