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
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2If you write the sum as 1/p_1 + ∑ (p_n)^-n (the sum being over n from 2 to infinity) then it is obviously less than 1/2 + ∑ (p_n)^-2 , and since the prime number set is a proper subset of the natural number set , the sum is obviously less than 1/2 + ζ(2) , whence convergent. – 2012-10-12
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1See also https://oeis.org/A093358 – 2012-10-22
1 Answers
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$$\sum_{n=1}^{\infty} \dfrac1{p_n^n} = \dfrac12 + \sum_{n=2}^{\infty} \dfrac1{p_n^n} \leq \dfrac12 + \sum_{n=2}^{\infty} \dfrac1{p_n^2} \leq \dfrac12 + \sum_{n=2}^{\infty} \dfrac1{n^2} = \dfrac12 + \dfrac{\pi^2}6 - 1 = \dfrac{\pi^2}6 - \dfrac12$$