Let $p_n$ denote the $n$ -th prime number. Evaluate the sum
$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{ \log p_n}{p_n^2 -1}$$
On a prime summation
On a prime summation
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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