1

In Apostol's analytic number theory book, a proof that $\sum \frac{1}{p_n}$, where $p_n$ represents the primes, diverges is given. In the proof, the number $Q = p_1p_2 \cdots p_k$ is formed, and the series $\sum_{n=1}^r \frac{1}{1+nQ}$ for $r \geq 1$ is considered. Since none of $p_j$ for $1 \leq j \leq k$ divides $1+nQ$, the prime factors must be in $p_{k+1}, p_{k+2}, \ldots$

Now, Apostol goes on to write $$ \sum_{n=1}^r \frac{1}{1+nQ} \leq \sum_{t=1}^\infty \left(\sum_{m=k+1}^\infty \frac{1}{p_m} \right)^t. $$

I don't understand how this is true. The only thing I can see here is $$ \sum_{n=1}^r \frac{1}{1+nQ} \leq \sum_{m=k+1}^\infty \frac{1}{p_m}. $$

Can someone please clarify this for me? Thanks in advance.

PS: He says that the sun on the right includes among its terms all the terms of the sum on the left. I don't see this.

IamRigour
  • 133
  • 2
    Does this answer your question: https://math.stackexchange.com/questions/361308/clarksons-proof-of-the-divergence-of-the-sum-of-the-reciprocals-of-the-primes?rq=1 – Philip Speegle Mar 05 '25 at 04:01
  • How exactly does $\frac{1}{1+nQ}$ appear in $$\left(\frac{1}{p_{k+1}} +\frac{1}{p_{k+2}} + \dots \right)^M = \frac{f(p_{k+1}, p_{k+2}, \dots )}{\prod_{m \geq k+1} p_m^M}$$ – IamRigour Mar 05 '25 at 04:29
  • $\frac1{1+nQ} = \frac1{p_{i_1}\cdots p_{i_M}}$ for some $M$ such that $p_{k+1}\le p_{i_j}$. Now expand the series on the left. That term will be in there. – Philip Speegle Mar 05 '25 at 04:36

0 Answers0