Odtnhj

Recall that the radical of an integer $n$ is defined to be $operatornamerad(n) = prod_p mid n p$.

For a paper, I need the result that
$$sum_n leq x frac1operatornamerad(n) ll_varepsilon x^varepsilon tag$*$,$$
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 $sum_nfrac1n^varepsilonrmrad(n)$. It suffices to show that it converges. However, it can be written as a product of
$$
S(p)=1+p^-1-varepsilon+p^-1-2varepsilon+dots=1+p^-1-varepsilonfrac 11-p^-varepsilonle 1+p^-1-fracvarepsilon 2
$$
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

$$sum_n:operatornamerad(n)=m frac1n^varepsilon=m^-varepsilonprod_pmid m(1-p^-varepsilon)^-1ll_varepsilon d(m)/m^varepsilon,$$

thus, the series

$$r(s)=sum_n=1^+infty frac1n^smathrmrad(n)$$

converges absolutely when $mathrmRe,s>0$. Now, using multiplicativity, one has

$$r(s)=prod_p (1+p^-s-1+p^-2s-1+ldots)=prod_p (1+frac1(1-p^-s)p^1+s).$$

Next, notice that for positive $varepsilon$ we have $1-2^-varepsilongg varepsilon$ and $1-p^-varepsilongeq varepsilon$ for $p>2$ and $varepsilon<1/6$. Therefore we deduce for any $varepsilon>0$

$$r(varepsilon)ll prod_pleft(1+frac1varepsilon p^1+varepsilonright)leq zeta(1+varepsilon)^1/varepsilon.$$

As $zeta(1+varepsilon)=frac1varepsilon+O(1)$, we finally obtain

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

Using Rankin trick we arrive at

$$sum_nleq x frac1mathrmrad(n)ll x^varepsilon varepsilon^-1/varepsilon.$$

Choosing $varepsilon=sqrtfraclnln x2ln x$ we prove that

$$sum_nleq x frac1mathrmrad(n)leq exp(sqrt(2+o(1))ln xlnln x),$$

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 $$sum_n le x frac1mathrmrad(n) = exp((1+o(1)) sqrt8logx/loglogx),$$ 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

$$displaystyle sum_n leq X frac1textrad(n) = sum_substackm leq X \ m text square-free frac1m sum_substackn leq X \ textrad(n) = m 1.$$

Now, $textrad(n) = m$ if and only if $p | n Rightarrow p | m$. If we write $m = p_1 cdots p_k$, then

$$displaystyle sum_substackn leq X \ textrad(n) = m 1 = #(x_1, cdots, x_k) : x_i in mathbbZ cap [0,infty), p_1^x_1 cdots p_k^x_k leq X/m.$$

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

$$displaystyle # (x_1, cdots, x_k) : x_1 log p_1 + cdots + x_k log p_k leq log(X/m) ll fraclog(X/m)prod_1 leq i leq k log(p_i) ll log X.$$

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

$$displaystyle # (x_1, cdots, x_k) : x_1 log p_1 + cdots + x_k log p_k leq log(X/m) = O left(sum_i=0^k frac(log X/m)^k-iprod_1 leq j leq k-i log p_i right).$$

It then follows that

$$displaystyle sum_n leq X frac1textrad(n) ll sum_substackp_1 < cdots < p_k \ p_1 cdots p_k leq X sum_i=0^k frac(log X)^k-iprod_1 leq j leq k -i p_i log p_i.$$

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.