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.
Best Answer
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 $\frac{1}{2}\ln n$.