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 2010-05-24, 22:38 #1 Damian     May 2005 Argentina 2728 Posts Sum of reciprocal of squares of all prime numbers It is known that both the armonic series and the sum of the reciprocals of the prime numbers diverges Informally written: $\sum_p \frac{1}{p} = \infty$ Considering that the sum of the reciprocal of all natural numbers converges, as seen in Basel problem, that is $\sum_n \frac{1}{n^2} = \frac{\pi^2}{6}$ I was wondering if the sum of the reciprocals of the squares of prime numbers converges, and if so to what number, that is $\sum_p \frac{1}{p^2} = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{11^2} + \ldots$ I tested numerically with Maxima software for the firsts primes with the code Code: sum (if primep(x) then 1/x^2 else 0, x, 2, 100000); and it seems to converge to some number that starts with 0.45224661779206... Any help is welcomed, thanks.
 2010-05-24, 22:53 #2 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 25×5×7 Posts It is easily seen to be convergent by a comparison test: http://mathworld.wolfram.com/ComparisonTest.html You are asking for the value of the prime zeta function at 2: http://en.wikipedia.org/wiki/Prime_zeta_function http://mathworld.wolfram.com/PrimeZetaFunction.html Wolfram gives references that may be helpful.
2010-05-24, 23:46   #3

"Richard B. Woods"
Aug 2002
Wisconsin USA

769210 Posts

Quote:
 Originally Posted by Damian the sum of the reciprocal of all natural numbers converges
You meant, "the sum of the reciprocals of the squares of all natural numbers converges". :-)

 2010-05-24, 23:57 #4 Damian     May 2005 Argentina 2×3×31 Posts Hi philmoore, thanks for your fast answer! I wasn't aware of the prime zeta function. So yes, my question was basically about $P(2)$. In the references it says that the firsts digits are $P(2) \simeq 0.452247420041065498506543364832247934173231343239892421736418 \ldots$ and that P. Sebah found more than 10000 digits. So only the firsts 5 decimals where correct on my original post. (And I guess a closed form isn't known for this number) Thanks again for your very useful answer.

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