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?
  • $\begingroup$ Both sums are not absolutely convergent. But the sum of the reciprocals of the primes grows extremely slow ! $\endgroup$
    – Peter
    Feb 13, 2023 at 14:22
  • 1
    $\begingroup$ 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 $\endgroup$
    – Henry
    Feb 13, 2023 at 14:22
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    $\begingroup$ It is known that $\sum_{p\le x}\frac{1}{p}\sim\log\log x$. $\endgroup$
    – Riemann
    Feb 13, 2023 at 14:23
  • $\begingroup$ @Henry Usually "conditionally convergent" implies that there is no absolute convergence. $\endgroup$
    – Peter
    Feb 13, 2023 at 14:27
  • $\begingroup$ 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 $\endgroup$
    – Tensor
    Feb 13, 2023 at 16:09


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