# Sum of the reciprocals of radicals

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?

• 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). – literature-searcher Nov 15 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 $$\prod_p S(p)\le C\prod_p(1+p^{-1-\frac\varepsilon 2})\le\sum_n n^{-1-\frac\varepsilon 2}<+\infty$$

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)

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

https://projecteuclid-org.proxy-remote.galib.uga.edu/euclid.ijm/1255631814

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).

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.

• I worry a bit about the uniformity in the "easy to see that" estimate in the $p_j$. Still, a very natural approach. – Greg Martin Nov 16 at 0:50
• 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 $\log{X}$. – so-called friend Don Nov 16 at 3:36
• The volume of the simplex should involve $(\log(X/m))^k$ rather than just $\log(X/m)$. This changes the bound dramatically. – Emil Jeřábek Nov 16 at 13:46
• 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 $\exp\left(\bigl(1+o(1)\bigr)\frac{\log X}{\log\log X}\right)$. – Emil Jeřábek Nov 16 at 14:58
• Now that I see it, this might begin to explain where Gerhard Paseman got his bound. – Emil Jeřábek Nov 16 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.

• 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. – Gerhard Paseman Nov 16 at 17:14