I recently saw this equation in a text book of mine and I am having a hard time following it. It shows how Euler's proof that the sum of the reciprocals of the primes diverge. This section deals with the fact that the original proof invokes real numbers and results from calculus, but then states that in principle a purely arithmetical proof can be given. It then proceeds to show that by starting with:
$$\sum_{i=0}^∞ \frac{1}{p_i}$$
Where the p's are primes. Then says suppose this sums converges then we can choose an N so large that:
$$\sum_{i=N}^∞ \frac{1}{p_i} < 1$$
Let A be the set containing 1 together with those positive integers all of whose prime factors are among $p_1,...,p_{N-1}$. And let B be the set containing 1 together with those positive integers all of whose prime factors are among the $p_i$for $i\le N$. We then have:
$$\sum_{n}^∞ \frac{1}{n} = \sum_{m\in A, k \in B} \frac{1}{mk} =(\sum_{m\in A} \frac{1}{m})(\sum_{k \in B} \frac{1}{k})$$
In this last product we can see that the first factor is bounded since it is a finite product from the above argument. We will then show that the second factor is bounded as well. From here the textbook loses me and I struggle to follow the logic. The textbook then says let $P(k_1,...,k_n,j_1,...,j_n,m)$ abbreviate the condition $k_1 + ... + k_n = m, k_1,...,k_n \gt0, N \le j_1 \lt j_2 \lt ... \lt j_n$. Then: $$\sum_{k\in B} \frac{1}{k} = \sum_{m} \sum_{P(k_1,...,k_n,j_1,...,j_n,m)} \frac{1}{p^{k_1}_{j_1}...p^{k_n}_{j_n}}= \sum_{m} (\sum_{j=N}^∞ \frac{1}{p_i})^m$$
The textbook then says:"The last sum (m summed over all natural numbers) converges by the condition on N. Thus we get that the harmonic series converges, and since this is not the case, it follows that $\sum_{i=0}^∞ \frac{1}{p_i}$ doesn't converge either". Though I understand the conclusion I struggle to follow how the sum of $\frac{1}{k}$ is equal to those terms on the right hand side. Thanks for any help!.