How to determine the convergence of $ \sum_{n=1}^\infty (p_n)^{-n}, $ where $p_n$ is the $n$th prime?
Convergence of $ \sum_{n=1}^\infty (p_n)^{-n}$
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sequences-and-series
prime-numbers
convergence-divergence
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1See also https://oeis.org/A093358 – 2012-10-22
1 Answers
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$\sum_{n=1}^{\infty} \dfrac1{p_n^n} = \dfrac12 + \sum_{n=2}^{\infty} \dfrac1{p_n^n} \leq \dfrac12 + \sum_{n=2}^{\infty} \dfrac1{p_n^2} \leq \dfrac12 + \sum_{n=2}^{\infty} \dfrac1{n^2} = \dfrac12 + \dfrac{\pi^2}6 - 1 = \dfrac{\pi^2}6 - \dfrac12$