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

If this result is true, can we prove convergence?