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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.

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    $\begingroup$ If you want to see something more obscure and impressive, see Ivan Niven's proof that $\pi$ is irrational in about 1/2 a page. More accessible (and elementary) is Paul Erdos' short proof of Berthold's "Postulate" (when Erdos was 18!), first proposed in the 1830's and first proved rigorously by Minkowski. $\endgroup$ Commented Feb 24, 2021 at 6:24
  • $\begingroup$ A Google search reveals nothing, perhaps you mean Bertrand's postulate? $\endgroup$ Commented Jun 2, 2021 at 13:13

4 Answers 4

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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 $$\prod_{p<n}\left(1+\frac{1}{p}\right)\geqslant\sum_{k<n}'\frac{1}{k} $$

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He explains his reasoning with respect to the first inequality in the paragraph above the inequality. Since each integer can be uniquely expressed as the product of a square-free integer and a square, each term on the right-hand sum occurs as a product of a single term from the square-free sum and a single term from the square sum. The inequality isn't strict because as $k$ and $j$ approach $n$, their product may exceed $n$, yielding some additional terms in the sum.

Similarly, in the second inequality, if you simply expand the product on the left, you will get all of the terms in the right-hand sum, and possibly some additional terms that you can throw away.

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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,

$$\prod_{p<n}\left(1+\frac1p\right)\ge\sum_{k<n}'\frac1k\,.$$

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For the first one, as Niven says every positive integer $m$ can be expressed uniquely in the form $m = kj^2$ where $k$ is squarefree. Obviously if $m<n$, then $k, j < n$ too. By distributivity, the left-hand side is $\sum_{k,j} 1/(kj^2)$ where $k, j < n$ and $k$ is squarefree, so every term on the right is included (along with a few more).

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