Prime Harmonic Series $\sum\limits_{p\in\mathbb P}\frac1p$ We have following identity: ($p$ is a prime number)
$$\left(1+\frac{1}{p}\right)\sum_{k=0}^n\frac{1}{p^{2k}}=\sum_{k=0}^{2n+1}\frac{1}{p^k}$$
Now, How to derive the following inequality from the above identity? Or use this identity to prove following inequality?
$$\prod_{p<N}\left(1+\frac{1}{p}\right)\sum_{k<N}\frac{1}{k^2}\ge\sum_{n<N}\frac{1}{n}.$$
This inequality shows that $\sum\frac{1}{p}$ is divergent.
 A: For a prime $p$ and an integer $n \neq 0$, let $v_p(n) = \max \{ k \geqslant 0 : p^k \mid n\}$. For $0 < m \leqslant N$, let ($\mathbb{P}$ is the set of (positive) primes)
$$S(m,N) := \bigl\{ k: 1 \leqslant k < N,\; \bigl(\forall p \in \mathbb{P}\bigr) \bigl(p \geqslant m \Rightarrow v_p(k) \equiv 0 \pmod{2}\bigr)\bigr\}.$$
$S(2,N)$ is the set of all squares $< N$, so
$$\sum_{n \in S(2,N)} \frac1n \leqslant \sum_{k<N} \frac{1}{k^2},$$
and $S(N,N) = \{n : 1 \leqslant n < N\}$. Now,
$$S(p_k,N) \cup p_k\cdot S(p_k,N) \supset S(p_{k+1},N),$$
since for $n \in S(p_{k+1},N)$, either $v_{p_k}(n) \equiv 0 \pmod{2}$, in which case $n \in S(p_k,N)$, or $v_{p_k}(n) \equiv 1 \pmod{2}$, in which case $n/p_k \in S(p_k,N)$. Hence, we have
$$\left(1 + \frac{1}{p_k}\right)\sum_{n \in S(p_k,N)} \frac1n = \sum_{n \in S(p_k,N) \cup p_k\cdot S(p_n,N)} \frac1n \geqslant \sum_{n\in S(p_{k+1},N)} \frac1n,$$
and therefore
$$\begin{align}
\prod_{j=1}^{\pi(N-1)}\left(1 +\frac{1}{p_j}\right)\sum_{k < N} \frac{1}{k^2} &\geqslant \prod_{j=1}^{\pi(N-1)}\left(1 +\frac{1}{p_j}\right)\sum_{n \in S(2,N)} \frac1n\\
&\geqslant \prod_{j=2}^{\pi(N-1)}\left(1 +\frac{1}{p_j}\right)\sum_{n \in S(p_2,N)} \frac1n\\
&\geqslant \dotsb\\
&\geqslant \sum_{n\in S(p_{\pi(N-1)+1},N)} \frac1n\\
&= \sum_{n < N} \frac1n.
\end{align}$$
