Clarkson's Proof of the Divergence of Reciprocal of Primes In Tom Apostol's book, he credits the proof of the divergence of the sum of reciprocal of primes to Clarkson. To begin, we assume $\{p_n\}$ is an enumeration of the primes and $$\sum_{n=1}^\infty\frac{1}{p_n}$$ is convergent. Then there exists a $k$ such that $$\sum_{n=k+1}^\infty\frac{1}{p_n}<\frac{1}{2}.$$ Defining $Q=p_1p_2\cdots p_k$, then, for each $r>0$, we get the inequality $$\sum_{n=1}^r \frac{1}{1+nQ}\leq\sum_{t=1}^\infty\left(\sum_{n=k+1}^\infty\frac{1}{p_n}\right)^t$$by noticing every term on the left appears on the right. This is where I have trouble, I know that all the prime factors of $1+nQ$ must be a subset of $\{p_n\}_{n=k+1}^\infty$, but I don't see how every term on the left appears on the right, can someone clarify this for me?
Also, I don't see how this LEADS to the infinitude of the primes. It seems we first must assume there are infinitely many in order to make this argument.
 A: From the definition of $Q$ it follows that each $1+nQ$ can only be divisible by primes larger than $p_k$. Thus we can write each
$$
\frac{1}{1+nQ} = \frac{1}{p_{i_1}p_{i_2}\cdots p_{i_M}}
$$
for some $M=M(n)$ where $k+1\le i_1,i_2,\ldots,i_M$ (and the $i$ are not necessarily distinct, allowing primes to appear more than once in the factorization). Then this term must appear in the expansion of
$$
\left(\frac{1}{p_{k+1}}+\frac{1}{p_{k+2}}+\frac{1}{p_{k+3}}+\cdots\right)^M
$$
Although this is written with the implicit assumption of the infinitude of primes, it is not necessary for the argument. We can just write
$$
\sum_{n=1}^r \frac{1}{1+nQ}\leq\sum_{t=1}^\infty\left(\sum_{n>k}\frac{1}{p_n}\right)^t
$$
allowing the set of remaining primes to be finite or infinite, and the result is the same.
In Apostol the infinitude of primes is established before this result. I don't think he intends to say that it follows from this proof, he only mentions in a historical note that Euler proved this result in 1737 (presumably by a different method) and noted the implication.
