Ivan Niven published a much shorter and easier proof that the prime harmonic series $\sum 1/p$ diverges. The proof is now available as open access here in this link.
In the short proof he makes use of two inequalities:
$$ \left( \sum_{k<n}'1/k \right) \left( \sum_{j<n}1/j^2 \right) \geq \sum_{m<n} 1/m$$
$k$ are the square-free integers, $j$ and $m$ are ordinary integers, and
$$\prod_{p<n}(1+1/p) \geq \sum_{k<n}'1/k$$
Question: Is there a simple logical rationale for why these inequalities hold?
I can only verify these inequalities by hand, multiplying out series for small $n$. Obviously this isn't proof, but doing this can sometimes reveal the more general logic. In this case I can't see it.
I would appreciate solutions and comments that avoid technical terminology as I, and my students, are not university trained mathematicians.
Update: Since Niven's article is now open access, here is a screenshot and a transcription:
$$\text{A PROOF OF THE DIVERGENCE OF $\Sigma 1 / p$}$$
$$\text{Ivan Niven, University of Oregon}$$
First we prove that $\Sigma^{\prime} 1 / k$ diverges, where $\Sigma^{\prime}$ denotes the sum over the squarefree positive integers. Each positive integer is uniquely expressible as a product of a squarefree positive integer and a square, so for any positive integer $n$,
$$
\left({\sum_{k<n}}^{\prime} 1 / k\right)\left(\sum_{j<n} 1 / j^{2}\right) \color{red}\geqq \sum_{m<n} 1 / m
$$
Here the second sum is bounded but the third sum is unbounded as $n$ increases, so the first sum must be unbounded. Next suppose that $\Sigma 1 / p$ converges to $\beta$, the sum taken over all primes $p$. By dropping all terms beyond $x$ in the series expansion of $e^{x}$ or $\exp (x),$ we see that $\exp (x)>1+x$ for $x>0 .$ Hence for each positive integer $n$
$$
\exp (\beta)>\exp \left(\sum_{p<n} 1 / p\right)=\prod_{p<n} \exp (1 / p)>\prod_{p<n}(1+1 / p) \color{red}\geqq {\sum_{k<n}}^{\prime} 1 / k
$$
But this contradicts the unboundedness of the last sum, so $\Sigma 1 / p$ diverges.
Best Answer
Each $m<n$ can be uniquely expressed as $m=kj^2$ where $k$ is a square-free integer, therefore $$ \sum_{m<n}\frac{1}{m}=\sum_{\substack{k,j \\ kj<n}}\frac{1}{kj^2}\leqslant\sum_{k,j<n}\frac{1}{kj^2}=\left(\sum_{k<n}'\frac{1}{k}\right)\left(\sum_{j<n}\frac{1}{j^2}\right) $$ where $k$ is a square free integer in the second and the third sum. As for the last inequality, let $p_1,\ldots,p_r$ be the prime numbers $<n$, then developping the product gives $$ \prod_{p<n}\left(1+\frac{1}{p}\right)=\sum_{I\subset[\![1,r]\!]}\prod_{i\in I}\frac{1}{p_i}$$ Each square-free $k<n$ has their prime divisors among $p_1,\ldots,p_r$ and thus is of the form $\prod_{i\in I}p_i$ where $I\subset[\![1,r]\!]$, therefore $$\prod_{p<n}\left(1+\frac{1}{p}\right)\geqslant\sum_{k<n}'\frac{1}{k} $$