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.


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.


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