Sum of the reciprocals of radicals
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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?
nt.number-theory reference-request
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up vote
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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?
nt.number-theory reference-request
4
The solutions given are what is sometimes called "Rankin's trick", that is, multiplying a series $nle X$ 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
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up vote
13
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favorite
up vote
13
down vote
favorite
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?
nt.number-theory reference-request
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?
nt.number-theory reference-request
nt.number-theory reference-request
edited Nov 15 at 23:01
Michael Hardy
5,53455383
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asked Nov 15 at 15:57
Daniel Loughran
10.8k22468
10.8k22468
4
The solutions given are what is sometimes called "Rankin's trick", that is, multiplying a series $nle X$ 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
add a comment |
4
The solutions given are what is sometimes called "Rankin's trick", that is, multiplying a series $nle X$ 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
4
4
The solutions given are what is sometimes called "Rankin's trick", that is, multiplying a series $nle X$ 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
The solutions given are what is sometimes called "Rankin's trick", that is, multiplying a series $nle X$ 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
add a comment |
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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 $prod_p S(p)le Cprod_p(1+p^{-1-fracvarepsilon 2})lesum_n n^{-1-fracvarepsilon 2}<+infty$
add a comment |
up vote
10
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First, notice that for any squarefree $m$ and any $varepsilon>0$ we have
notice that
$$sum_{n:operatorname{rad}(n)=m} frac{1}{n^varepsilon}=m^{-varepsilon}prod_{pmid m}(1-p^{-varepsilon})^{-1}ll_varepsilon d(m)/m^varepsilon,$$
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)
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8
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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
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).
3
Wolfgang Schwarz refined de Bruijn's results at the Tauberian level, but I think it's still a log asymptotic (his ${mathcal R}$-function is delicate to deal with IIRC). He had three papers on the general subject, the second of which is the most relevant (I give all 3 links). digizeitschriften.de/dms/img/?PID=GDZPPN002181304 digizeitschriften.de/dms/img/?PID=GDZPPN002181339 digizeitschriften.de/en/dms/img/?PID=GDZPPN002182629
– literature-searcher
Nov 16 at 4:12
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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.
4
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
3
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
1
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 $sum_{nle X}frac1{mathrm{rad}(n)}leprod_{ple X}left(1+frac{log X}{plog p}right)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)frac{log X}{loglog X}right)$.
– 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
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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.
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
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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 $prod_p S(p)le Cprod_p(1+p^{-1-fracvarepsilon 2})lesum_n n^{-1-fracvarepsilon 2}<+infty$
add a comment |
up vote
13
down vote
accepted
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 $prod_p S(p)le Cprod_p(1+p^{-1-fracvarepsilon 2})lesum_n n^{-1-fracvarepsilon 2}<+infty$
add a comment |
up vote
13
down vote
accepted
up vote
13
down vote
accepted
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 $prod_p S(p)le Cprod_p(1+p^{-1-fracvarepsilon 2})lesum_n n^{-1-fracvarepsilon 2}<+infty$
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 $prod_p S(p)le Cprod_p(1+p^{-1-fracvarepsilon 2})lesum_n n^{-1-fracvarepsilon 2}<+infty$
answered Nov 15 at 17:05
fedja
36.6k7105201
36.6k7105201
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First, notice that for any squarefree $m$ and any $varepsilon>0$ we have
notice that
$$sum_{n:operatorname{rad}(n)=m} frac{1}{n^varepsilon}=m^{-varepsilon}prod_{pmid m}(1-p^{-varepsilon})^{-1}ll_varepsilon d(m)/m^varepsilon,$$
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)
add a comment |
up vote
10
down vote
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)
add a comment |
up vote
10
down vote
up vote
10
down vote
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)
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)
edited Nov 16 at 12:58
answered Nov 15 at 16:59
Asymptotiac K
1,2241313
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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
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).
3
Wolfgang Schwarz refined de Bruijn's results at the Tauberian level, but I think it's still a log asymptotic (his ${mathcal R}$-function is delicate to deal with IIRC). He had three papers on the general subject, the second of which is the most relevant (I give all 3 links). digizeitschriften.de/dms/img/?PID=GDZPPN002181304 digizeitschriften.de/dms/img/?PID=GDZPPN002181339 digizeitschriften.de/en/dms/img/?PID=GDZPPN002182629
– literature-searcher
Nov 16 at 4:12
add a comment |
up vote
8
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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
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).
3
Wolfgang Schwarz refined de Bruijn's results at the Tauberian level, but I think it's still a log asymptotic (his ${mathcal R}$-function is delicate to deal with IIRC). He had three papers on the general subject, the second of which is the most relevant (I give all 3 links). digizeitschriften.de/dms/img/?PID=GDZPPN002181304 digizeitschriften.de/dms/img/?PID=GDZPPN002181339 digizeitschriften.de/en/dms/img/?PID=GDZPPN002182629
– literature-searcher
Nov 16 at 4:12
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up vote
8
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up vote
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down vote
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).
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).
answered Nov 16 at 3:22
so-called friend Don
4,97811720
4,97811720
3
Wolfgang Schwarz refined de Bruijn's results at the Tauberian level, but I think it's still a log asymptotic (his ${mathcal R}$-function is delicate to deal with IIRC). He had three papers on the general subject, the second of which is the most relevant (I give all 3 links). digizeitschriften.de/dms/img/?PID=GDZPPN002181304 digizeitschriften.de/dms/img/?PID=GDZPPN002181339 digizeitschriften.de/en/dms/img/?PID=GDZPPN002182629
– literature-searcher
Nov 16 at 4:12
add a comment |
3
Wolfgang Schwarz refined de Bruijn's results at the Tauberian level, but I think it's still a log asymptotic (his ${mathcal R}$-function is delicate to deal with IIRC). He had three papers on the general subject, the second of which is the most relevant (I give all 3 links). digizeitschriften.de/dms/img/?PID=GDZPPN002181304 digizeitschriften.de/dms/img/?PID=GDZPPN002181339 digizeitschriften.de/en/dms/img/?PID=GDZPPN002182629
– literature-searcher
Nov 16 at 4:12
3
3
Wolfgang Schwarz refined de Bruijn's results at the Tauberian level, but I think it's still a log asymptotic (his ${mathcal R}$-function is delicate to deal with IIRC). He had three papers on the general subject, the second of which is the most relevant (I give all 3 links). digizeitschriften.de/dms/img/?PID=GDZPPN002181304 digizeitschriften.de/dms/img/?PID=GDZPPN002181339 digizeitschriften.de/en/dms/img/?PID=GDZPPN002182629
– literature-searcher
Nov 16 at 4:12
Wolfgang Schwarz refined de Bruijn's results at the Tauberian level, but I think it's still a log asymptotic (his ${mathcal R}$-function is delicate to deal with IIRC). He had three papers on the general subject, the second of which is the most relevant (I give all 3 links). digizeitschriften.de/dms/img/?PID=GDZPPN002181304 digizeitschriften.de/dms/img/?PID=GDZPPN002181339 digizeitschriften.de/en/dms/img/?PID=GDZPPN002182629
– literature-searcher
Nov 16 at 4:12
add a comment |
up vote
2
down vote
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.
4
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
3
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
1
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 $sum_{nle X}frac1{mathrm{rad}(n)}leprod_{ple X}left(1+frac{log X}{plog p}right)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)frac{log X}{loglog X}right)$.
– 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
add a comment |
up vote
2
down vote
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.
4
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
3
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
1
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 $sum_{nle X}frac1{mathrm{rad}(n)}leprod_{ple X}left(1+frac{log X}{plog p}right)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)frac{log X}{loglog X}right)$.
– 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
add a comment |
up vote
2
down vote
up vote
2
down vote
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.
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.
edited Nov 16 at 16:40
answered Nov 16 at 0:49
Stanley Yao Xiao
8,27942783
8,27942783
4
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
3
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
1
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 $sum_{nle X}frac1{mathrm{rad}(n)}leprod_{ple X}left(1+frac{log X}{plog p}right)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)frac{log X}{loglog X}right)$.
– 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
add a comment |
4
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
3
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
1
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 $sum_{nle X}frac1{mathrm{rad}(n)}leprod_{ple X}left(1+frac{log X}{plog p}right)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)frac{log X}{loglog X}right)$.
– 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
4
4
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
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
3
3
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
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
1
1
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
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 $sum_{nle X}frac1{mathrm{rad}(n)}leprod_{ple X}left(1+frac{log X}{plog p}right)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)frac{log X}{loglog X}right)$.
– Emil Jeřábek
Nov 16 at 14:58
Using a correct formula for the volume, I get $sum_{nle X}frac1{mathrm{rad}(n)}leprod_{ple X}left(1+frac{log X}{plog p}right)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)frac{log X}{loglog X}right)$.
– 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
Now that I see it, this might begin to explain where Gerhard Paseman got his bound.
– Emil Jeřábek
Nov 16 at 15:05
add a comment |
up vote
2
down vote
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.
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
add a comment |
up vote
2
down vote
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.
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
add a comment |
up vote
2
down vote
up vote
2
down vote
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.
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.
edited Nov 16 at 18:51
answered Nov 15 at 19:58
Gerhard Paseman
8,19411845
8,19411845
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
add a comment |
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
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
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
add a comment |
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The solutions given are what is sometimes called "Rankin's trick", that is, multiplying a series $nle X$ 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