5
$\begingroup$

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.

$\endgroup$
12
  • $\begingroup$ Looks like you are taking a divergent series minus an absolute convergent series...that should be obvious... $\endgroup$
    – imranfat
    Jul 29 '16 at 21:36
  • $\begingroup$ Yes but I was asking for verification of the correctness of the proof. $\endgroup$
    – Legendre
    Jul 29 '16 at 21:38
  • 1
    $\begingroup$ Thanks. But more specifically, we are adding what seems to be an infinite number of convergent series, namely the ones that are of the form c * 1/n^2. How can we be sure that this still converges? $\endgroup$
    – Legendre
    Jul 29 '16 at 21:40
  • 4
    $\begingroup$ The sum of not squarefrees contains $\sum \frac1{4n}$, hence diverges $\endgroup$ Jul 29 '16 at 21:55
  • 1
    $\begingroup$ That's not the point. The sum I am referring to is in the long run less than about $\ln n$ times $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots$. This is $\frac{\pi^2}{6}-1$, which is nicely less than $1$. We don't need to know the exact sum $\sum_1^\infty \frac{1}{k^2}$, a reasonably good upper bound will do. So the sum of the reciprocals of square-frees up to $n$ is bounded below by a non-zero constant times $\ln n$. $\endgroup$ Jul 29 '16 at 22:21
2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ $(\ln n) (\frac 1{3\cdot 4} + \cdots )$ is a lower bound of the infinite sum right above that. $\endgroup$ Jul 29 '16 at 23:10
  • $\begingroup$ @i707107: Thanks, I was missing the first two terms, separated out for the estimate. $\endgroup$ Jul 29 '16 at 23:25
2
$\begingroup$

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 $\log x \rightarrow\infty$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.