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### Sum of Inverse of Primes

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Date: 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?
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```
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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Associated Topics:
High School Sequences, Series

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