Sum of the reciprocals of radicals
Clash Royale CLAN TAG#URR8PPP
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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?
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 $operatornamerad(n) = prod_p mid n p$.
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?
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
2 days ago
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up vote
12
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up vote
12
down vote
favorite
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?
nt.number-theory reference-request
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?
nt.number-theory reference-request
nt.number-theory reference-request
edited 2 days ago
Michael Hardy
5,53455383
5,53455383
asked 2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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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 $prod_p S(p)le Cprod_p(1+p^-1-fracvarepsilon 2)lesum_n n^-1-fracvarepsilon 2<+infty$
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
$$sum_n:operatornamerad(n)=m frac1n^varepsilon=m^-varepsilonprod_pmid m(1-p^-varepsilon)^-1ll_varepsilon d(m)/m^varepsilon,$$
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)
1
It would be good for you to edit this answer so that it is correct.
– Lucia
yesterday
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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 $$sum_n le x frac1mathrmrad(n) = exp((1+o(1)) sqrt8logx/loglogx),$$ 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).
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
yesterday
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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
$$displaystyle sum_n leq X frac1textrad(n) = sum_substackm leq X \ m text square-free frac1m sum_substackn leq X \ textrad(n) = m 1.$$
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.
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
yesterday
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 $logX$.
– so-called friend Don
yesterday
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
yesterday
Using a correct formula for the volume, I get $sum_nle Xfrac1mathrmrad(n)leprod_ple Xleft(1+fraclog Xplog pright)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)fraclog Xloglog Xright)$.
– Emil Jeřábek
yesterday
Now that I see it, this might begin to explain where Gerhard Paseman got his bound.
– Emil Jeřábek
yesterday
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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
yesterday
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5 Answers
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up vote
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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 $prod_p S(p)le Cprod_p(1+p^-1-fracvarepsilon 2)lesum_n n^-1-fracvarepsilon 2<+infty$
add a comment |
up vote
12
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accepted
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 $prod_p S(p)le Cprod_p(1+p^-1-fracvarepsilon 2)lesum_n n^-1-fracvarepsilon 2<+infty$
add a comment |
up vote
12
down vote
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up vote
12
down vote
accepted
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 $prod_p S(p)le Cprod_p(1+p^-1-fracvarepsilon 2)lesum_n n^-1-fracvarepsilon 2<+infty$
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 $prod_p S(p)le Cprod_p(1+p^-1-fracvarepsilon 2)lesum_n n^-1-fracvarepsilon 2<+infty$
answered 2 days ago
fedja
36.5k7105200
36.5k7105200
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up vote
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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)
1
It would be good for you to edit this answer so that it is correct.
– Lucia
yesterday
add a comment |
up vote
10
down vote
First, notice that for any squarefree $m$ and any $varepsilon>0$ we have
notice that
$$sum_n:operatornamerad(n)=m frac1n^varepsilon=m^-varepsilonprod_pmid m(1-p^-varepsilon)^-1ll_varepsilon d(m)/m^varepsilon,$$
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)
1
It would be good for you to edit this answer so that it is correct.
– Lucia
yesterday
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
$$sum_n:operatornamerad(n)=m frac1n^varepsilon=m^-varepsilonprod_pmid m(1-p^-varepsilon)^-1ll_varepsilon d(m)/m^varepsilon,$$
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)
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)
edited yesterday
answered 2 days ago
Asymptotiac K
1,2241313
1,2241313
1
It would be good for you to edit this answer so that it is correct.
– Lucia
yesterday
add a comment |
1
It would be good for you to edit this answer so that it is correct.
– Lucia
yesterday
1
1
It would be good for you to edit this answer so that it is correct.
– Lucia
yesterday
It would be good for you to edit this answer so that it is correct.
– Lucia
yesterday
add a comment |
up vote
8
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 $$sum_n le x frac1mathrmrad(n) = exp((1+o(1)) sqrt8logx/loglogx),$$ 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).
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
yesterday
add a comment |
up vote
8
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 $$sum_n le x frac1mathrmrad(n) = exp((1+o(1)) sqrt8logx/loglogx),$$ 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).
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
yesterday
add a comment |
up vote
8
down vote
up vote
8
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 $$sum_n le x frac1mathrmrad(n) = exp((1+o(1)) sqrt8logx/loglogx),$$ 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).
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).
answered yesterday
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
yesterday
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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
yesterday
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
yesterday
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
yesterday
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
$$displaystyle sum_n leq X frac1textrad(n) = sum_substackm leq X \ m text square-free frac1m sum_substackn leq X \ textrad(n) = m 1.$$
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.
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
yesterday
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 $logX$.
– so-called friend Don
yesterday
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
yesterday
Using a correct formula for the volume, I get $sum_nle Xfrac1mathrmrad(n)leprod_ple Xleft(1+fraclog Xplog pright)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)fraclog Xloglog Xright)$.
– Emil Jeřábek
yesterday
Now that I see it, this might begin to explain where Gerhard Paseman got his bound.
– Emil Jeřábek
yesterday
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
$$displaystyle sum_n leq X frac1textrad(n) = sum_substackm leq X \ m text square-free frac1m sum_substackn leq X \ textrad(n) = m 1.$$
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.
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
yesterday
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 $logX$.
– so-called friend Don
yesterday
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
yesterday
Using a correct formula for the volume, I get $sum_nle Xfrac1mathrmrad(n)leprod_ple Xleft(1+fraclog Xplog pright)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)fraclog Xloglog Xright)$.
– Emil Jeřábek
yesterday
Now that I see it, this might begin to explain where Gerhard Paseman got his bound.
– Emil Jeřábek
yesterday
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
$$displaystyle sum_n leq X frac1textrad(n) = sum_substackm leq X \ m text square-free frac1m sum_substackn leq X \ textrad(n) = m 1.$$
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.
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.
edited yesterday
answered yesterday
Stanley Yao Xiao
8,26442783
8,26442783
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
yesterday
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 $logX$.
– so-called friend Don
yesterday
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
yesterday
Using a correct formula for the volume, I get $sum_nle Xfrac1mathrmrad(n)leprod_ple Xleft(1+fraclog Xplog pright)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)fraclog Xloglog Xright)$.
– Emil Jeřábek
yesterday
Now that I see it, this might begin to explain where Gerhard Paseman got his bound.
– Emil Jeřábek
yesterday
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
yesterday
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 $logX$.
– so-called friend Don
yesterday
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
yesterday
Using a correct formula for the volume, I get $sum_nle Xfrac1mathrmrad(n)leprod_ple Xleft(1+fraclog Xplog pright)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)fraclog Xloglog Xright)$.
– Emil Jeřábek
yesterday
Now that I see it, this might begin to explain where Gerhard Paseman got his bound.
– Emil Jeřábek
yesterday
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
yesterday
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
yesterday
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 $logX$.
– so-called friend Don
yesterday
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 $logX$.
– so-called friend Don
yesterday
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
yesterday
The volume of the simplex should involve $(log(X/m))^k$ rather than just $log(X/m)$. This changes the bound dramatically.
– Emil Jeřábek
yesterday
Using a correct formula for the volume, I get $sum_nle Xfrac1mathrmrad(n)leprod_ple Xleft(1+fraclog Xplog pright)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)fraclog Xloglog Xright)$.
– Emil Jeřábek
yesterday
Using a correct formula for the volume, I get $sum_nle Xfrac1mathrmrad(n)leprod_ple Xleft(1+fraclog Xplog pright)$, which I believe can be bounded by $expleft(bigl(1+o(1)bigr)fraclog Xloglog Xright)$.
– Emil Jeřábek
yesterday
Now that I see it, this might begin to explain where Gerhard Paseman got his bound.
– Emil Jeřábek
yesterday
Now that I see it, this might begin to explain where Gerhard Paseman got his bound.
– Emil Jeřábek
yesterday
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
yesterday
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
yesterday
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 yesterday
answered 2 days ago
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
yesterday
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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
yesterday
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
yesterday
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
yesterday
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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
2 days ago