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Old 2010-05-24, 22:38   #1
Damian
 
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May 2005
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Default 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.
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Old 2010-05-24, 22:53   #2
philmoore
 
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"Phil"
Sep 2002
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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.
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Old 2010-05-24, 23:46   #3
cheesehead
 
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"Richard B. Woods"
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Quote:
Originally Posted by Damian View Post
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". :-)
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Old 2010-05-24, 23:57   #4
Damian
 
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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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