Clearing up proof of: Harmonic series of primes diverges 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...   

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.
