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Let us consider the series $\sum_{p\in\Bbb P}{\frac{\log{p}}{p^2-1}}$ where $p$ ranges through prime numbers. We can, using Euler product of Riemann zeta function prove that this series converges to $-\frac{\zeta^\prime(2)}{\zeta(2)}.$ I would like to know whether the limit of series $\sum_{p\in\Bbb P}{\frac{\log{p}}{(p^2-1)^2}}$ where $p$ ranges through prime numbers can be expressed using some values of Riemann zeta function.

Sorry if the question is not written as per community's standards. It is my first question here. Thank you.

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    $\begingroup$ Use $p$ instead of $l$ , its commonly used to denote the set of primes. $\endgroup$ – Ahmad Sep 8 '17 at 14:27
  • $\begingroup$ Expand $\frac{1}{(x-1)^2}$ in power series to obtain $$\sum_p \frac{\log p}{(p^2-1)^2} = -\sum_{k \ge 1} \frac{\zeta'(2k)}{\zeta(2k)} \sum_{d | k} \frac{\mu(d)}{d} $$ $\endgroup$ – reuns Sep 9 '17 at 22:19
  • $\begingroup$ @Ahmad $p$ is perfectly fine for primes just as $n$ is perfectly fine for natural numbers. There is no reason to suggest using $l$ for primes here as $p$ is not already used to denote something else. $\endgroup$ – Klangen Dec 17 '18 at 9:32

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