# Is alternating sum of reciprocal of primes absolutely convergent?

I remember reading that alternating harmonic function is a conditionally convergent series

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \frac{1}{1} - \frac{1}{2}+\frac{1}{3} - \frac{1}{4}+...$$

I wonder, whether alternating sum of reciprocals of primes is also a conditionally convergent series

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{p_n} = \frac{1}{2} - \frac{1}{3}+\frac{1}{5} - \frac{1}{7}+...$$

If it is, how do we prove it?
• Both sums are not absolutely convergent. But the sum of the reciprocals of the primes grows extremely slow ! Feb 13, 2023 at 14:22
• If the signs alternate and the magnitudes decline towards $0$ then the series conditionally converges. Just consider the partial sums after an odd number of terms and after an even number of terms Feb 13, 2023 at 14:22
• It is known that $\sum_{p\le x}\frac{1}{p}\sim\log\log x$. Feb 13, 2023 at 14:23
• @Henry Usually "conditionally convergent" implies that there is no absolute convergence. Feb 13, 2023 at 14:27
• How do we even proof that both series are conditionally convergent. Because, I think for sure that $$\sum_{n=0}^\infty (- \frac{1}{2})^{n+1}$$ is absolutely convergent Feb 13, 2023 at 16:09