Bdtyktl

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?

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

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

notice that

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

thus, the series

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

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 $varepsilon>0$

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

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

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

Using Rankin trick we arrive at

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

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

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

which is a bit non-optimal by the answer of Don. (But at least we have the correct $lnln$ 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 $$sum_{n le x} frac{1}{mathrm{rad}(n)} = exp((1+o(1)) sqrt{8log{x}/loglog{x}}),$$ as $xtoinfty$. 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.

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 $(e-1)p_0 + pi(p_0)loglog x$. We want this last quantity to be asymptotically less than $epsilonlog x$.

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 $( (log x)/f(x))/(loglog x - log f(x))$, so the second term is only slightly bigger than $log(x)/f(x)$, but small enough to dip below $epsilonlog x$.

If you put in some work, you find $f(x)$ is less than but close to $loglog x$, and far enough away for the fraction $(loglog x)/(loglog x - log f(x))$ 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.