Divergence of sum of reciprocals of square-free numbers

Question

I am aware that there is a similar question here, however I want to prove that $\sum \frac{1}{n}$ for $n$ square free diverges, without relying on the fact that $\sum \frac{1}{p}$ diverges for $p$ prime. This is equivalent to proving that $\sum \frac{|\mu(n)|}{n}$ converges, where $\mu(n)$ is the mobius function. I would like verification that my proof is correct:

My Proof: We begin by noting that $\sum_{n = 1}^{\infty} \frac{1}{n^2}$ converges, and so $$c * \sum_{n = 1}^{\infty} \frac{1}{n^2}$$ must also converge for any positive integer $c$. Therefore, if we look at the sum $\sum \frac{1}{n}$ where $n$ ranges only through the integers that are not squarefree, then this sum must converge because it is the composition of a series of convergent sums. Since $\sum_{n = 1}^{\infty} \frac{1}{n}$ diverges, we must have the sum of the reciprocals of square-free integers also diverging.

In particular, how can I be sure that the sum of the reciprocals of non square-free integers converges? It seems like we are taking an infinite number of convergent sums, which doesn't have to necessarily converge.

Answer 1

Outline: Let $n$ be large. Note that the sum of the reciprocals of the integers up to $n$ is about $\ln n$.

The sum of the reciprocals of multiples of $4$ up to $n$ is less than roughly $\frac{1}{4}\ln n$. Here we are already giving away a bit, since this reciprocal sum is actually about $\frac{1}{4}\ln(n/4)$.

The sum of the reciprocals of multiples of $9$ up to $n$ is less than roughly $\frac{1}{9}\ln n$. And so on.

So the sum of the reciprocals of not square-frees up to $n$ is less than roughly $$(\ln n)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots.\right).$$ The infinite sum above is upper bounded by $$(\ln n)\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\cdots\right),$$ which is less than $\frac{1}{2}\ln n$. (Since we are upper-bounding, we don't need to worry about overlap.)

So the sum of the reciprocals of the square-frees up to $n$ is asymptotically greater than $\frac{1}{2}\ln n$.

Answer 2

Let $Q(x)$ be the number of square-free positive integers $n\leq x$. It is well-known that $$ Q(x) = \frac 6{\pi^2} x + O(\sqrt x). $$ We now apply partial summation: $$ \sum_{n\leq x} \frac{\mu^2(n)}{n} = \frac {Q(t)}t \bigg\vert_{1-}^x + \int_1^x \frac{Q(t)}{t^2} dt = \frac 6{\pi^2} \log x+O(1). $$ Thus, the sum of reciprocals of square-free numbers diverges, as $\log x \rightarrow\infty$.

Answer 3

André Nicolas's proof is impeccable, but I'd like to give another one, which is probably closer to yours.

Let's recall that a family $(a_\lambda)_{\lambda \in \Lambda}$ of nonnegative numbers is summable is the set of finite sums $\left\{ \sum_{\lambda \in \Lambda'} a_\lambda \,\middle|\, \Lambda' \subset \Lambda\text{ finite}\right\}$ is bounded. In that case, the least upper bound of this set is the sum of the family, and we denote it $\sum_{\lambda\in\Lambda} a_\lambda$. This theory is basically equivalent to the theory of series (which corresponds to $\Lambda = \mathbb N$) if $\Lambda$ is countable, which is the only interesting case.

The result I'd like to use is the following, which corresponds to the Cauchy product of ordinary series.

Theorem. Let $(a_{\lambda})_{\lambda \in\Lambda}$ and $(b_\mu)_{\mu\in M}$ be two families of nonnegative numbers. If these two families are summable, then so is $(a_\lambda\,b_\mu)_{(\lambda, \mu)_\in\Lambda\times M}$, and we have the equality $$\sum_{(\lambda,\mu)\in \Lambda \times M} a_\lambda\,b_\mu = \left(\sum_{\lambda\in\Lambda} a_\lambda\right) \, \left(\sum_{\mu\in M} b_\mu\right).$$

Now, let $S = \{1, 4, 9, 16,\ldots\}$ and $SF = \{1, 2, 3, 5, 6, 7, 10,\ldots\}$ be the sets of square and squarefree numbers, respectively.

You already know that $(n^{-1})_{n \in\mathbb N^*}$ isn't summable and that $(n^{-1})_{n\in S}$ is (because that's really equivalent to saying that $\sum \frac 1{m^2}$ converges). I claim that the above theorem directly shows that these two properties imply that $(n^{-1})_{n\in SF}$ isn't summable.

Indeed, if $(n^{-1})_{n\in SF}$ were summable, so would $(a^{-1}\,b^{-1})_{(a,b)\in S\times SF}$, as per the theorem. But it is easy to show that every positive number can be written in a unique way as the product of a square number and a squarefree number. So, this product family $(a^{-1}\,b^{-1})_{(a,b)\in S\times SF}$ is really the same thing as $(n^{-1})_{n\in\mathbb N^*}$, so it cannot be summable, which proves that $(n^{-1})_{n\in SF}$ wasn't summable either.