TITLE: unexplained inequalities from Niven's 1971 proof $\sum 1/p$ diverges
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: When you multiply out the product
$$\left(\sum_{k<n}'\frac1k\right)\left(\sum_{j<n}\frac1{j^2}\right)\,,\tag{1}$$
you get the sum of all possible terms of the form $\frac1{kj^2}$, where $k$ is a square-free integer less than $n$, and $j$ is any integer less than $n$. Every positive integer $m<n$ can be expressed as the product of such a $k$ and $j^2$, so these product terms include every fraction $\frac1m$ for $m<n$. In general, however, they will also include some fractions $\frac1m$ with $m\ge n$: if $n-1$ is square-free, for instance, we can take $k=j=n-1$ and get $\frac1{(n-1)^3}$. Thus, every term of $\sum_{m<n}\frac1m$ appears in $(1)$ after the latter is multiplied out, and $(1)$ generally contains extra terms as well, so
$$\left(\sum_{k<n}'\frac1k\right)\left(\sum_{j<n}\frac1{j^2}\right)\ge\sum_{m<n}\frac1m\,.$$
For the second inequality let $P=\{p_1,\ldots,p_\ell\}$ be the set of primes less than $n$. When the product
$$\prod_{p<n}\left(1+\frac1p\right)=\prod_{i=1}^\ell\left(1+\frac1{p_i}\right)\tag{2}$$
is multiplied out, there is one term for each possible choice of one factor, $1$ or $\frac1{p_i}$, from each of the $\ell$ factors. Specifically, if $S\subseteq P$, one of the terms of the product is
$$\frac1{\prod_{i\in S}p_i}\,,\tag{3}$$
obtained by choosing $1$ when $i\notin S$ and $\frac1{p_i}$ when $i\in S$.
Every square-free integer less than $n$ is a product of distinct primes less than $n$, so every term of $\sum\limits_{k<n}'\frac1k$ is one of the terms $(3)$. However, there may also be some $S\subseteq P$ such that $\prod_{i\in S}p_i\ge n$, so the terms of $(2)$ after it’s multiplied out include all of the terms of $\sum\limits_{k<n}'\frac1k$ and possible some others as well. Thus,
$$\prod_{p<n}\left(1+\frac1p\right)\ge\sum_{k<n}'\frac1k\,.$$