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In the book "An Introduction to Number Theory" by Apostol there is a proof given by Clarkson, that $ \sum_{n=1}^{\infty} {1 \over {p_m}} $ diverges.

It is assumed $ \sum_{m=k+1}^{\infty} {1 \over {p_m}} < {1 \over 2} $ for a certain $ k \in \mathbb{N} $ . Then it defined $ Q $ as: $ Q = \prod_{i=1}^{m} p_m $

And reaches the equation: $ \sum_{n=1}^{r} {1 \over 1+n*Q} \le \sum_{t=1}^{\infty} (\sum_{m=k+1}^{\infty} {1 \over p_m}) ^t $, which is perfectly fine.

Now it is claimed that the "right side includes among its terms all the terms of the left." This I don't understand.

Well, it is obvious for $ n = 1 $ that $ n*Q+1 $ is a new prime and thus included on the right side. However I don't see how the quoted statement above holds, when $1+Q*n$ is not prime.

Though I find it clear, that all prime factors of $1+n*Q$ appear in the sum of the right side.

I would be very happy, if someone could clear up this misunderstanding, as the problem seems rather basic and I still fail to get it right...

Gottfried Helms
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Imago
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2 Answers2

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As has been pointed out, $Q+1$ need not be prime. But it's clear that none of $p_1,\dots,p_m$ can divide $1+n*Q$, since those primes all divide $Q$. So every $1+n*Q$ is a product of primes $p_j$ for $j>m$, and the reciprocal of any such product is one of the terms on the right.

David C. Ullrich
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  • The right side just covers the prime numbers. For example I choose 2 as the certain value of k. Therefore on the left side I have all the reciprocals of 7,13,19,25,... etc, in the proof it is the stated: A term of the left side is also on the right side. So for example: How can 1/55 be split into a sum of reciprocals of primes? – Imago Sep 06 '15 at 20:55
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    @Imago why don't you try writing the first few terms, with $t=1$ and $t=2$? Already $t=2$ will answer your question about 1/55. – guest Sep 06 '15 at 20:59
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    @Imago No. Look again. The right side is a sum of powers of sums. The first term, $t=1$, is just the sum of reciprocals of primes. The second term $t=2$ contains reciprocals or products of two primes... – David C. Ullrich Sep 06 '15 at 21:08
  • Oh! Haha.. yes... this was easy... thank you... I wonder how I can be so blind and narrow sighted.. – Imago Sep 06 '15 at 21:10
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For $n>0$ the number $nQ+1$ is a product of powers of primes that are all greater than $p_m$. The terms in the multinomial expansion of the RHS are reciprocals of products of powers of primes that are $>p_m$, and includes every such one at least once (Exactly once,actually,because prime decomposition is unique). So $1/(nQ+1)$ occurs as a RHS term for $ n=1...r$..... This is a cumbersome variant of Euler's proof : If $p_1,...,p_m$ are the only primes,then the set $S$ of all products of non-negative powers of $p_1,...,p_m $is $Z^+$. But $$\infty > M=\prod_{j=1}^{j=m} (1-1/p_j)^{-1} = \prod_{j=1}^{j=m} \sum_{k=0}^{k= \infty}p_j^{-k}$$ and multiplying out the RHS , the term $1/t$ occurs on thr RHS for every $t \in S$, so $$M \ge \sum_{t \in S} 1/t =\sum_{n=1}^{\infty}1/n=\infty$$ a contradiction.

DanielWainfleet
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