My Math Forum Reciprocal sum of primes diverges... proof?

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 December 27th, 2006, 10:01 PM #1 Newbie   Joined: Nov 2006 Posts: 20 Thanks: 0 Reciprocal sum of primes diverges... proof? We'll need the following definitions to solve this problem: definition: For any number x, Nj(x) is the number of numbers less than or equal to x whose only prime divisors are in the set of the first j primes { p1, p2, ..., pj }. definition: A number is square-free if none of its divisors is a perfect square (except 1). (a) Prove that there are exactly 2j square-free numbers whose only prime divisors are in the set { p1, p2, ..., pj }. (b) Prove that, for sufficiently large x, Nj(x)<= 2j * x (Hint: show every number can be written uniquely as the product of a perfect square and a square-free number). (c) Prove that, for sufficiently large x, Nj(x) < x. Why does this mean there must be an infinite number of primes? (d) For subsequent results, we need a little bit more. Prove that, for sufficiently large x, Nj(x) < x2 (The proof is exactly like the one in (c).) We will use (d) to prove: theorem: The sum of the reciprocals of the primes diverge! That is, 1p1 + 1p2 +1p3 +... diverges [Is it clear that this is even stronger than merely being an infinite set? For example, the perfect squares are an infinite set, but the sum of the reciprocals of the perfect squares converges.] Anyhow to show that the sum of the reciprocals of the primes diverges, it's enough to show that, for any j: 1/pj+1 + 1/pj+2 + 1/pj+3 + .... > 1/2 (e) Why is it enough to show that the above sum is greater than 12 ? (f) Show that for any x, x/pj+1 + x/pj+2 + x/pj+3 + .... >= x - Nj(x). (Hint: Explain first why x/p is greater than or equal to the number of numbers less than or equal to x that are divisible by p. What about x/p + x/q where p and q are relatively prime?) (g) Use the results from (d) and (f) to complete the proof that the sums of the reciprocals of the primes diverge.
 December 28th, 2006, 02:17 AM #2 Newbie   Joined: Nov 2006 Posts: 20 Thanks: 0 I have already solved part a), if anyone has idea in solving part b, part c, and part d, please do teach me. Thank you very much.
 December 28th, 2006, 02:36 PM #3 Newbie   Joined: Dec 2006 Posts: 29 Thanks: 0 For part a), I believe you meant 2^j. In part b), I'm seeing Nj(x)<= 2j * ?x, what was this supposed to say?
 December 28th, 2006, 06:36 PM #4 Newbie   Joined: Dec 2006 From: Moscow, Russia Posts: 4 Thanks: 0 b) Just use hint. Every n<=x can be uniquely written in the form n=a*b, where a is squrefree and b is perfect square. Number of possible values a is not more than 2^j (part (a)), number of possible values b is not more than sqrt(x) (because b=c^2 and c<=sqrt(x)). Hence number of possible values for n=ab is not more than 2^j*sqrt(x). c) If there were only finitely many prime numbers, namely {p1,p2,...,pj}, then for all integer x>0 we'd have Nj(x)=x. Contradiction. d) Obvious. (Does x2 mean x/2?) e) Use Cauchy criterion. f)x-Nj(x) equals quantity of numbers n<=x which have at least one prime factor from {p_{j+1}, p_{j+2},... } . Quantity of n<=x which are divisible by p is not more than x/p (because n=p*m and m<=x/p).
 December 29th, 2006, 05:45 AM #5 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 I asked the same question ( If the sum of the reciprocals of the prime numbers diverged ) on the AoPS forum. Here's the link, but we can discuss it here as well, of course. I believe it was Euler who proved this conjecture to be true. ( Euler proved everything ) http://www.artofproblemsolving.com/F...c.php?t=122342
 December 30th, 2006, 02:05 PM #6 Newbie   Joined: Nov 2006 Posts: 20 Thanks: 0 Hi infinity, Thank you very much for your reply and information. I found this link which is related to my question: http://primes.utm.edu/infinity.shtml With the help of this link, I got the idea to almost finish the question.

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