I'm trying to evaluate:

$\displaystyle{\sum_{k=1}^{\infty}\frac{1}{p_k(p_k-1)}}$, where $p_k$ is the $k$-th prime number.

But I cannot even figure out how to begin. I have a feeling that this could involve the prime zeta function, but I'm not sure. In fact, I'm not even sure an analytic closed-form solution is possible. Could you please help?


We can rewrite this as $\sum \limits_{k=1}^\infty ( \frac{1}{p_k^2}+\frac{1}{p_k^3}+\dots)$.

So we want $\sum\limits_{k=2}^\infty P(k)$ where $P$ is the prime zeta function.

  • 1
    $\begingroup$ $ = \sum_{k=2}^\infty \sum_{n=1}^\infty \frac{\mu(n)}{n}\log \zeta( n k) = \sum_{m=2}^\infty \log \zeta( m) \sum_{n | m,n \ne m } \frac{\mu(n)}{n} =\sum_{m=2}^\infty \log \zeta( m) \frac{\varphi(m)-\mu(m)}{m}$ $\endgroup$ – reuns Dec 28 '16 at 5:30

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