Recall that the radical of an integer $n$ is defined to be $\operatorname{rad}(n) = \prod_{p \mid n } p$.

For a paper, I need the result that $$\sum_{n \leq x} \frac{1}{\operatorname{rad}(n)} \ll_\varepsilon x^{\varepsilon} \tag{$*$},$$ for all $\varepsilon > 0$. I have a proof of this using complex analysis and Perron's formula, but this seems a bit overkill given that I'm looking for a weak upper bound for a problem in elementary number theory.

Does anyone know of a short elementary proof of the bound $(*)$? Or better yet, a reference?

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    $\begingroup$ The solutions given are what is sometimes called "Rankin's trick", that is, multiplying a series $n\le X$ by $(X/n)^\alpha$ and optimizing $\alpha$. My recollection is that getting an asymptotic for your sum is rather difficult (though again IIRC a log asymptotic is viable by the saddlepoint method). $\endgroup$ – literature-searcher Nov 15 '18 at 19:03

You can get away with elementary analytic number theory. Consider the series $\sum_n\frac{1}{n^{\varepsilon}\rm{rad}(n)}$. It suffices to show that it converges. However, it can be written as a product of $$ S(p)=1+p^{-1-\varepsilon}+p^{-1-2\varepsilon}+\dots=1+p^{-1-\varepsilon}\frac 1{1-p^{-\varepsilon}}\le 1+p^{-1-\frac\varepsilon 2} $$ for all but finitely many $p$. Thus $\prod_p S(p)\le C\prod_p(1+p^{-1-\frac\varepsilon 2})\le\sum_n n^{-1-\frac\varepsilon 2}<+\infty$

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First, notice that for any squarefree $m$ and any $\varepsilon>0$ we have

notice that

$$\sum_{n:\operatorname{rad}(n)=m} \frac{1}{n^\varepsilon}=m^{-\varepsilon}\prod_{p\mid m}(1-p^{-\varepsilon})^{-1}\ll_\varepsilon d(m)/m^\varepsilon,$$

thus, the series

$$r(s)=\sum_{n=1}^{+\infty} \frac{1}{n^s\mathrm{rad}(n)}$$

converges absolutely when $\mathrm{Re}\,s>0$. Now, using multiplicativity, one has

$$r(s)=\prod_p (1+p^{-s-1}+p^{-2s-1}+\ldots)=\prod_p (1+\frac{1}{(1-p^{-s})p^{1+s}}).$$

Next, notice that for positive $\varepsilon$ we have $1-2^{-\varepsilon}\gg \varepsilon$ and $1-p^{-\varepsilon}\geq \varepsilon$ for $p>2$ and $\varepsilon<1/6$. Therefore we deduce for any $\varepsilon>0$

$$r(\varepsilon)\ll \prod_p\left(1+\frac{1}{\varepsilon p^{1+\varepsilon}}\right)\leq \zeta(1+\varepsilon)^{1/\varepsilon}.$$

As $\zeta(1+\varepsilon)=\frac{1}{\varepsilon}+O(1)$, we finally obtain

$$r(\varepsilon)\ll \varepsilon^{-1/\varepsilon}.$$

Using Rankin trick we arrive at

$$\sum_{n\leq x} \frac{1}{\mathrm{rad}(n)}\ll x^\varepsilon \varepsilon^{-1/\varepsilon}.$$

Choosing $\varepsilon=\sqrt{\frac{\ln\ln x}{2\ln x}}$ we prove that

$$\sum_{n\leq x} \frac{1}{\mathrm{rad}(n)}\leq \exp(\sqrt{(2+o(1))\ln x\ln\ln x}),$$

which is a bit non-optimal by the answer of Don. (But at least we have the correct $\ln\ln$ asymptotics)

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de Bruijn studies this sum in "On the number of integers $\le x$ whose prime factors divide $n$", which was published in a 1962 volume of the Illinois J. Math; see


He proves there (see Theorem 1) that $$\sum_{n \le x} \frac{1}{\mathrm{rad}(n)} = \exp((1+o(1)) \sqrt{8\log{x}/\log\log{x}}),$$ as $x\to\infty$. Of course, this implies the $O(x^{\epsilon})$ bound you were after. However, his proof (which uses a Tauberian theorem of Hardy and Ramanujan) is not as elementary as some others that have been suggested here (but gives a more precise result).

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sIt seems that this argument hasn't been presented yet, so I might as well include it.

We can sort the integers $n \in [1, X]$ by their radicals, which is necessarily a square-free integer $m$. Thus we have

$$\displaystyle \sum_{n \leq X} \frac{1}{\text{rad}(n)} = \sum_{\substack{m \leq X \\ m \text{ square-free}}} \frac{1}{m} \sum_{\substack{n \leq X \\ \text{rad}(n) = m}} 1.$$

Now, $\text{rad}(n) = m$ if and only if $p | n \Rightarrow p | m$. If we write $m = p_1 \cdots p_k$, then

$$\displaystyle \sum_{\substack{n \leq X \\ \text{rad}(n) = m}} 1 = \#\{(x_1, \cdots, x_k) : x_i \in \mathbb{Z} \cap [0,\infty), p_1^{x_1} \cdots p_k^{x_k} \leq X/m\}.$$

The inequality defining the right hand side is equivalent to

$$\displaystyle x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m),$$

and this is just counting integer points with non-negative entries bounded by a simplex, and it is easy to see that

$$\displaystyle \# \{(x_1, \cdots, x_k) : x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m)\} \ll \frac{\log(X/m)}{\prod_{1 \leq i \leq k} \log(p_i)} \ll \log X.$$

EDIT: This last step is wrong, but it can be fixed. Indeed, we can arrange the $p_i$'s so that $p_1 < p_2 < \cdots < p_k$. It then follows from Davenport's lemma that

$$\displaystyle \# \{(x_1, \cdots, x_k) : x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m)\} = O \left(\sum_{i=0}^k \frac{(\log X/m)^{k-i}}{\prod_{1 \leq j \leq k-i} \log p_i} \right).$$

It then follows that

$$\displaystyle \sum_{n \leq X} \frac{1}{\text{rad}(n)} \ll \sum_{\substack{p_1 < \cdots < p_k \\ p_1 \cdots p_k \leq X}} \sum_{i=0}^k \frac{(\log X)^{k-i}}{\prod_{1 \leq j \leq k -i} p_i \log p_i}.$$

From here I think it is possible to get the bound $O_\epsilon(X^\epsilon)$, but it requires a somewhat more refined analysis on the interaction between the number of primes and the size of the primes.

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    $\begingroup$ I worry a bit about the uniformity in the "easy to see that" estimate in the $p_j$. Still, a very natural approach. $\endgroup$ – Greg Martin Nov 16 '18 at 0:50
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    $\begingroup$ Note that de Bruijn's estimate (quoted in my answer) shows that Greg's concern is a serious one: the sum is in fact not bounded by any fixed power of $\log{X}$. $\endgroup$ – so-called friend Don Nov 16 '18 at 3:36
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    $\begingroup$ The volume of the simplex should involve $(\log(X/m))^k$ rather than just $\log(X/m)$. This changes the bound dramatically. $\endgroup$ – Emil Jeřábek Nov 16 '18 at 13:46
  • $\begingroup$ Using a correct formula for the volume, I get $\sum_{n\le X}\frac1{\mathrm{rad}(n)}\le\prod_{p\le X}\left(1+\frac{\log X}{p\log p}\right)$, which I believe can be bounded by $\exp\left(\bigl(1+o(1)\bigr)\frac{\log X}{\log\log X}\right)$. $\endgroup$ – Emil Jeřábek Nov 16 '18 at 14:58
  • $\begingroup$ Now that I see it, this might begin to explain where Gerhard Paseman got his bound. $\endgroup$ – Emil Jeřábek Nov 16 '18 at 15:05

Here is another approach. Let $p_0$ be the largest prime with $(p_0)^{(e-1)p_0} \leq x$. The desired sum is bounded above by $P =\prod_{p}(1+\lfloor \log_p x \rfloor/p)$, where the product is over primes $p$ less than or equal to $x$.

When we pick out those terms of $P$ whose numerator is $k$, and consider the product of just those terms, we look at those primes with $p^k \lt x \leq p^{k+1}$ and the log of that product is bounded by $k$ times the sum $ S_k$ of $1/p$ over those primes. Mertens theorem gives $\log((k+1)/k)$ as an approximate value for $S_k$ for small $k$, so the subproduct is approximated by $((k+1)/k)^k$. So for $k=1$ up to just before $(e-1)p_0$, we have broken the product over larger primes than $p_0$ into sub products each bounded by $e$.

So we have an immediate upper bound on $P$ of $(1 + (\log_2 x)/2)^{\pi(p_0)}e^{(e-1)p_0}$. For $x$ not too small, this is less than $(\log x)^{\pi(p_0)}e^{(e-1)p_0}$. So far we have log of your sum is dominated by $\log P$ which in turn is dominated by $(e-1)p_0 + \pi(p_0)\log\log x$. We want this last quantity to be asymptotically less than $\epsilon\log x$.

Well, $(e-1)p_0 \leq (\log x)/(\log p_0)$, so $p_0 \lt (\log x)/f(x)$ for a function $f(x)$ which is slowly increasing. But $\pi(p_0)$ is asymptotically $( (\log x)/f(x))/(\log\log x - \log f(x))$, so the second term is only slightly bigger than $\log(x)/f(x)$, but small enough to dip below $\epsilon\log x$.

If you put in some work, you find $f(x)$ is less than but close to $\log\log x$, and far enough away for the fraction $(\log\log x)/(\log\log x - \log f(x))$ not to be a problem. Although the prime number theorem and Mertens theorem on sum 1/p are used, this should be elementary enough.

Observation 2018.11.16 Since a weak result is wanted, we can weaken some of the requirements: replace the prime number theorem by a result that bounds $\pi(p)$ from above by $Ap/\log p$ , and regroup the terms of the partial product $P$ into pieces each of which multiply to a number less than $e^2$. One should not need the full strength of Mertens for this. Or, follow the suggestion in the comment below and focus on the product of the biggest $\pi(p_0)$ terms, and show the difference between this product and the sum is sufficiently small. End Observation 2018.11.16.

Gerhard "For Some Value Of 'Enough'" Paseman, 2018.11.15.

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  • $\begingroup$ One can also upper bound the sum by dividing it into two: one with terms where the radical includes primes bigger than p_0, and one with terms where the radical has no primes bigger than p_0. The argument above shows that the first part is less substantial than the second part. This suggests to me that looking at the second part (sum over p_0-smooth numbers) is more interesting and requires more delicacy. Gerhard "Waves Hands Over Hard Parts" Paseman, 2018.11.16. $\endgroup$ – Gerhard Paseman Nov 16 '18 at 17:14

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