Sum of Inverse of PrimesDate: 05/25/99 at 12:48:43 From: Manuel Puertas Subject: Sum of primes' inverse We have the infinite series: S = 1/1 + 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + ... 1/p(n) + ... where p(n) = the nth prime. Is S convergent or divergent? Why? Date: 05/25/99 at 16:26:33 From: Doctor Rob Subject: Re: Sum of primes' inverse Thanks for writing to Ask Dr. Math. S diverges. The partial sums SUM 1/p for p <= x grow like log(log(x)). Here is a proof from Hardy and Wright, _The Theory of Numbers_, 4th edition, p. 17: First let N(x) be the number of n <= x and not divisible by any prime p > p(j). Then any such n has the form n = a^2*m, where m is squarefree. Now m = 2^b(1) * 3^b(2) * ... * p(j)^b(j), where each b(i) is 0 or 1, so there are at most 2^j possible values of m, and at most sqrt(x) possible values of a, so (*) N(x) <= 2^j*sqrt(x). Now if the series is convergent, we can choose j so that the remainder after j terms is less than 1/2, i.e., 1/p(j+1) + 1/p(j+2) + ... < 1/2. The number of n <= x which are divisible by p is at most x/p. Hence x - N(x), the number of n <= x divisible by one or more of p(j+1), p(j+2), ..., is not more than x/p(j+1) + x/p(j+2) + ... < x/2. Hence by equation (*) above, x/2 < N(x) <= 2^j*sqrt(x), so x < 2^(2*j+2), which is false for x >= 2^(2*j+2). Thus the series diverges. Q.E.D. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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