About the limit $lim_n to +infty frac1n^2 sum_1 le a,b le n frac1 mathrmgcd (a,b) $Interesting phenomenon with the $zeta(3)$ seriesTo what extent are divisibility by different primes independent?System of congruences that do not satisfy CRT assumptions (via algorithm)Is the limit of a Dirichlet series to infinity always equal to the constant term?Practical probability that the GCD of a set of large, random integers is 1?About the differentiability of $sum_n=1^inftyfracmu(n)sin(sqrt2n^3x)sqrt2n^3e^-isqrtn^3x$On a limit involving $sum_n=1^inftyfracsin(sqrtn^3+1x)sqrtn^3+1$On the equation $(sigma(ab))^2=2a^2(sigma(b))^2+2b^2(sigma(a))^2$, for integers $1<a<b$Estimate median of Cauchy distributionAnomaly in probability of 3 randomly-selected integers being coprime
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About the limit $lim_n to +infty frac1n^2 sum_1 le a,b le n frac1 mathrmgcd (a,b) $
Interesting phenomenon with the $zeta(3)$ seriesTo what extent are divisibility by different primes independent?System of congruences that do not satisfy CRT assumptions (via algorithm)Is the limit of a Dirichlet series to infinity always equal to the constant term?Practical probability that the GCD of a set of large, random integers is 1?About the differentiability of $sum_n=1^inftyfracmu(n)sin(sqrt2n^3x)sqrt2n^3e^-isqrtn^3x$On a limit involving $sum_n=1^inftyfracsin(sqrtn^3+1x)sqrtn^3+1$On the equation $(sigma(ab))^2=2a^2(sigma(b))^2+2b^2(sigma(a))^2$, for integers $1<a<b$Estimate median of Cauchy distributionAnomaly in probability of 3 randomly-selected integers being coprime
$begingroup$
This is not homework.
My question is:
Prove or disprove: $$lim_n to +infty frac1n^2 sum_a,b=1^n frac1 mathrmgcd (a,b) = fraczeta(3)zeta(2)$$
This would represent the probability as $n to +infty$ that, after having picked randomly $a,b,c in 1, dots ,n $, the line
$$ax+by+c=0$$
has some integer coordinate point $(x,y in Bbb Z^2)$. Indeed, fixed $a,b in 1, dots ,n $, then $c=-ax-by$ for some $x,y in Bbb Z$ is equivalent to $$c mathrm is a multiple of gcd(a,b)$$
because of Bezout identity; this event happens with probability $1/gcd (a,b)$.
What I tried first: clearly
$$0 le frac1n^2 sum_a,b=1^n frac1 mathrmgcd (a,b) le frac1n^2 sum_a,b=1^n 1 = 1$$
so, our limit belongs to $[0,1]$ (if it exists).
Then I made some euristic argument:
I know that the probability, as $n to +infty$, that two random numbers $a,bin 1, dots ,n $ are coprime is $$zeta(2)^-1 = frac6pi^2,$$ thus the probability that $gcd(a,b)=d$ is equal to
$$fraczeta(2)^-1d^2 = frac6d^2pi^2$$
Thus our probability is (and here is where I have doubts in my argument)
$$sum_d=1^n fraczeta(2)^-1d^2 to frac1zeta(2) cdot zeta(2) =1$$
EDIT: or maybe my last step should be
$$sum_d=1^n fraczeta(2)^-1d^2 frac1d to fraczeta(3)zeta(2)$$
number-theory limits greatest-common-divisor probability-limit-theorems
asked Mar 17 at 10:57
CrostulCrostul
28.2k22352
$endgroup$
add a comment |
$begingroup$
This is not homework.
My question is:
Prove or disprove: $$lim_n to +infty frac1n^2 sum_a,b=1^n frac1 mathrmgcd (a,b) = fraczeta(3)zeta(2)$$
This would represent the probability as $n to +infty$ that, after having picked randomly $a,b,c in 1, dots ,n $, the line
$$ax+by+c=0$$
has some integer coordinate point $(x,y in Bbb Z^2)$. Indeed, fixed $a,b in 1, dots ,n $, then $c=-ax-by$ for some $x,y in Bbb Z$ is equivalent to $$c mathrm is a multiple of gcd(a,b)$$
because of Bezout identity; this event happens with probability $1/gcd (a,b)$.
What I tried first: clearly
$$0 le frac1n^2 sum_a,b=1^n frac1 mathrmgcd (a,b) le frac1n^2 sum_a,b=1^n 1 = 1$$
so, our limit belongs to $[0,1]$ (if it exists).
Then I made some euristic argument:
I know that the probability, as $n to +infty$, that two random numbers $a,bin 1, dots ,n $ are coprime is $$zeta(2)^-1 = frac6pi^2,$$ thus the probability that $gcd(a,b)=d$ is equal to
$$fraczeta(2)^-1d^2 = frac6d^2pi^2$$
Thus our probability is (and here is where I have doubts in my argument)
$$sum_d=1^n fraczeta(2)^-1d^2 to frac1zeta(2) cdot zeta(2) =1$$
EDIT: or maybe my last step should be
$$sum_d=1^n fraczeta(2)^-1d^2 frac1d to fraczeta(3)zeta(2)$$
number-theory limits greatest-common-divisor probability-limit-theorems
asked Mar 17 at 10:57
CrostulCrostul
28.2k22352
$endgroup$
2
$begingroup$
The limit cannot be $1$ since for each $n$, there are at least $frac(n-1)^24$ pairs of even numbers $a,b$ with $textgcd(a,b)ge 2$. So the limit is at most $1cdot frac 3 4+frac 12 frac 14 =frac 7 8.$
$endgroup$
– Song
Mar 17 at 11:32
add a comment |
$begingroup$
This is not homework.
My question is:
Prove or disprove: $$lim_n to +infty frac1n^2 sum_a,b=1^n frac1 mathrmgcd (a,b) = fraczeta(3)zeta(2)$$
This would represent the probability as $n to +infty$ that, after having picked randomly $a,b,c in 1, dots ,n $, the line
$$ax+by+c=0$$
has some integer coordinate point $(x,y in Bbb Z^2)$. Indeed, fixed $a,b in 1, dots ,n $, then $c=-ax-by$ for some $x,y in Bbb Z$ is equivalent to $$c mathrm is a multiple of gcd(a,b)$$
because of Bezout identity; this event happens with probability $1/gcd (a,b)$.
What I tried first: clearly
$$0 le frac1n^2 sum_a,b=1^n frac1 mathrmgcd (a,b) le frac1n^2 sum_a,b=1^n 1 = 1$$
so, our limit belongs to $[0,1]$ (if it exists).
Then I made some euristic argument:
I know that the probability, as $n to +infty$, that two random numbers $a,bin 1, dots ,n $ are coprime is $$zeta(2)^-1 = frac6pi^2,$$ thus the probability that $gcd(a,b)=d$ is equal to
$$fraczeta(2)^-1d^2 = frac6d^2pi^2$$
Thus our probability is (and here is where I have doubts in my argument)
$$sum_d=1^n fraczeta(2)^-1d^2 to frac1zeta(2) cdot zeta(2) =1$$
EDIT: or maybe my last step should be
$$sum_d=1^n fraczeta(2)^-1d^2 frac1d to fraczeta(3)zeta(2)$$
number-theory limits greatest-common-divisor probability-limit-theorems
asked Mar 17 at 10:57
CrostulCrostul
28.2k22352
$endgroup$
This is not homework.
My question is:
Prove or disprove: $$lim_n to +infty frac1n^2 sum_a,b=1^n frac1 mathrmgcd (a,b) = fraczeta(3)zeta(2)$$
This would represent the probability as $n to +infty$ that, after having picked randomly $a,b,c in 1, dots ,n $, the line
$$ax+by+c=0$$
has some integer coordinate point $(x,y in Bbb Z^2)$. Indeed, fixed $a,b in 1, dots ,n $, then $c=-ax-by$ for some $x,y in Bbb Z$ is equivalent to $$c mathrm is a multiple of gcd(a,b)$$
because of Bezout identity; this event happens with probability $1/gcd (a,b)$.
What I tried first: clearly
$$0 le frac1n^2 sum_a,b=1^n frac1 mathrmgcd (a,b) le frac1n^2 sum_a,b=1^n 1 = 1$$
so, our limit belongs to $[0,1]$ (if it exists).
Then I made some euristic argument:
I know that the probability, as $n to +infty$, that two random numbers $a,bin 1, dots ,n $ are coprime is $$zeta(2)^-1 = frac6pi^2,$$ thus the probability that $gcd(a,b)=d$ is equal to
$$fraczeta(2)^-1d^2 = frac6d^2pi^2$$
Thus our probability is (and here is where I have doubts in my argument)
$$sum_d=1^n fraczeta(2)^-1d^2 to frac1zeta(2) cdot zeta(2) =1$$
EDIT: or maybe my last step should be
$$sum_d=1^n fraczeta(2)^-1d^2 frac1d to fraczeta(3)zeta(2)$$
number-theory limits greatest-common-divisor probability-limit-theorems
number-theory limits greatest-common-divisor probability-limit-theorems
asked Mar 17 at 10:57
CrostulCrostul
28.2k22352
asked Mar 17 at 10:57
CrostulCrostul
28.2k22352
edited Mar 17 at 11:45
Crostul
asked Mar 17 at 10:57
CrostulCrostul
28.2k22352
asked Mar 17 at 10:57
CrostulCrostul
28.2k22352
asked Mar 17 at 10:57
CrostulCrostul
28.2k22352
28.2k22352
2
$begingroup$
The limit cannot be $1$ since for each $n$, there are at least $frac(n-1)^24$ pairs of even numbers $a,b$ with $textgcd(a,b)ge 2$. So the limit is at most $1cdot frac 3 4+frac 12 frac 14 =frac 7 8.$
$endgroup$
– Song
Mar 17 at 11:32
add a comment |
2
$begingroup$
The limit cannot be $1$ since for each $n$, there are at least $frac(n-1)^24$ pairs of even numbers $a,b$ with $textgcd(a,b)ge 2$. So the limit is at most $1cdot frac 3 4+frac 12 frac 14 =frac 7 8.$
$endgroup$
– Song
Mar 17 at 11:32
2
2
$begingroup$
The limit cannot be $1$ since for each $n$, there are at least $frac(n-1)^24$ pairs of even numbers $a,b$ with $textgcd(a,b)ge 2$. So the limit is at most $1cdot frac 3 4+frac 12 frac 14 =frac 7 8.$
$endgroup$
– Song
Mar 17 at 11:32
$begingroup$
The limit cannot be $1$ since for each $n$, there are at least $frac(n-1)^24$ pairs of even numbers $a,b$ with $textgcd(a,b)ge 2$. So the limit is at most $1cdot frac 3 4+frac 12 frac 14 =frac 7 8.$
$endgroup$
– Song
Mar 17 at 11:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is no need to involve probabilities, just a bit of analytic number theory suffices. I don't know how familiar you are with $L$-functions, so I'll leave some details open for now. If any part needs clarification, let me know.
For every positive integer $n$ define
$$f(n):=sum_a,b=1^nfrac1gcd(a,b)
qquadtext and qquad
g(n):=sum_dmid nfracvarphi(tfracnd)d$$ so that for all $n>1$ we have
begineqnarray*
f(n)-f(n-1)
&=&-frac1n+2sum_c=1^nfrac1gcd(c,n)\
&=&-frac1n+2sum_dmid nfracvarphi(tfracnd)d\
&=&-frac1n+2g(n).
endeqnarray*
Because $lim_ntoinftyfrac1n^2sum_k=1^n-frac1k=0$, this already shows that
$$lim_ntoinftyfrac1n^2f(n)=2lim_ntoinftyfrac1n^2sum_k=1^ng(n).$$
Note that $g$ is a multiplicative function, and in fact $g=iotaastvarphi$ where $iota(n):=tfrac1n$ and $ast$ denotes the Dirichlet convolution of multiplicative functions. This immediately shows that the $L$-function associated to $g$ converges for all $sinBbbC$ with $operatornameRe(s)>2$ and that for all such $sinBbbC$ we have
$$L_g(s)=L_iota(s)L_varphi(s)=zeta(s+1)L_varphi(s).$$
It is easy to see that $L_iota(s)=zeta(s+1)$ for these $s$, and it is a basic exercise to show that
$$L_varphi(s)=fraczeta(s-1)zeta(s),$$
whenever $operatornameRe(s)>2$. Then by a Tauberian theorem, for example Wiener-Ikehara, we have
begineqnarray*
lim_ntoinftyfrac1n^2f(n)
&=&2operatornameRes(L_g,2)\
&=&2cdotoperatornameRes(zeta(s+1),2)cdotoperatornameRes(L_varphi,2)\
&=&2cdotzeta(3)cdotfrac12zeta(2)\
&=&fraczeta(3)zeta(2).
endeqnarray*
answered Mar 19 at 21:00
ServaesServaes
29.9k342101
$endgroup$
$begingroup$
Not familiar with $L$-functions at all. Anyway, thanks for your answer.
$endgroup$
– Crostul
Mar 19 at 23:25
$begingroup$
the point is that $L_g(s) = zeta(s+1)zeta(s-1)/zeta(s)$ then $L_g(s) = sum_nge 1 g(n) n^-s=sint_1^infty sum_n le x g(n) x^-s-1dx$ implies $sum_n le x g(n) = frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ and the Tauberian theorem is about using analyticity and boundedness to shift $int_3-iinfty^3+iinfty$ to the left, same method as the one developed for the PNT @Crostul
$endgroup$
– reuns
Mar 29 at 0:31
$begingroup$
@reuns I have included a mention of this, I won't provide a proof here.
$endgroup$
– Servaes
Mar 29 at 0:35
$begingroup$
The main difficulty is that $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ doesn't converge absolutely, we can look instead at $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ss^2ds$ or $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^s+1s(s+1)ds$ then deduce the asymptotic for the original one
$endgroup$
– reuns
Mar 29 at 0:37
$begingroup$
@reuns I agree that there are some subtle details in applying the theorem, but seeing that the OP is not familiar with $L$-functions at all, I thought I'd omit those until asked for.
$endgroup$
– Servaes
Mar 29 at 0:39
add a comment |
$begingroup$
Your doubts are justified: $sum_d=1^n fraczeta(2)^-1d^2$ is just a sum of limiting probabilities, so it should (and does) yield $1$ in the limit. The sum you want to compute is of values of $frac1gcd(a,b)$, so you need to weight the possible values of that expression by their probabilities. This gives
$$
sum_d=1^n fraczeta(2)^-1d^2cdotfrac1dtofraczeta(3)zeta(2)approx 0.730763
$$
A numerical experimentat (with $n=1000$) gives a result close to $0.731$, so this seems likely to be correct.
answered Mar 17 at 11:18
FredHFredH
3,6851023
$endgroup$
$begingroup$
Thank you for your answer! Indeed I was having doubts while writing my question. Anyway, this is just heuristics, now what I'm looking for is a proof.
$endgroup$
– Crostul
Mar 17 at 11:23
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is no need to involve probabilities, just a bit of analytic number theory suffices. I don't know how familiar you are with $L$-functions, so I'll leave some details open for now. If any part needs clarification, let me know.
For every positive integer $n$ define
$$f(n):=sum_a,b=1^nfrac1gcd(a,b)
qquadtext and qquad
g(n):=sum_dmid nfracvarphi(tfracnd)d$$ so that for all $n>1$ we have
begineqnarray*
f(n)-f(n-1)
&=&-frac1n+2sum_c=1^nfrac1gcd(c,n)\
&=&-frac1n+2sum_dmid nfracvarphi(tfracnd)d\
&=&-frac1n+2g(n).
endeqnarray*
Because $lim_ntoinftyfrac1n^2sum_k=1^n-frac1k=0$, this already shows that
$$lim_ntoinftyfrac1n^2f(n)=2lim_ntoinftyfrac1n^2sum_k=1^ng(n).$$
Note that $g$ is a multiplicative function, and in fact $g=iotaastvarphi$ where $iota(n):=tfrac1n$ and $ast$ denotes the Dirichlet convolution of multiplicative functions. This immediately shows that the $L$-function associated to $g$ converges for all $sinBbbC$ with $operatornameRe(s)>2$ and that for all such $sinBbbC$ we have
$$L_g(s)=L_iota(s)L_varphi(s)=zeta(s+1)L_varphi(s).$$
It is easy to see that $L_iota(s)=zeta(s+1)$ for these $s$, and it is a basic exercise to show that
$$L_varphi(s)=fraczeta(s-1)zeta(s),$$
whenever $operatornameRe(s)>2$. Then by a Tauberian theorem, for example Wiener-Ikehara, we have
begineqnarray*
lim_ntoinftyfrac1n^2f(n)
&=&2operatornameRes(L_g,2)\
&=&2cdotoperatornameRes(zeta(s+1),2)cdotoperatornameRes(L_varphi,2)\
&=&2cdotzeta(3)cdotfrac12zeta(2)\
&=&fraczeta(3)zeta(2).
endeqnarray*
answered Mar 19 at 21:00
ServaesServaes
29.9k342101
$endgroup$
$begingroup$
Not familiar with $L$-functions at all. Anyway, thanks for your answer.
$endgroup$
– Crostul
Mar 19 at 23:25
$begingroup$
the point is that $L_g(s) = zeta(s+1)zeta(s-1)/zeta(s)$ then $L_g(s) = sum_nge 1 g(n) n^-s=sint_1^infty sum_n le x g(n) x^-s-1dx$ implies $sum_n le x g(n) = frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ and the Tauberian theorem is about using analyticity and boundedness to shift $int_3-iinfty^3+iinfty$ to the left, same method as the one developed for the PNT @Crostul
$endgroup$
– reuns
Mar 29 at 0:31
$begingroup$
@reuns I have included a mention of this, I won't provide a proof here.
$endgroup$
– Servaes
Mar 29 at 0:35
$begingroup$
The main difficulty is that $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ doesn't converge absolutely, we can look instead at $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ss^2ds$ or $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^s+1s(s+1)ds$ then deduce the asymptotic for the original one
$endgroup$
– reuns
Mar 29 at 0:37
$begingroup$
@reuns I agree that there are some subtle details in applying the theorem, but seeing that the OP is not familiar with $L$-functions at all, I thought I'd omit those until asked for.
$endgroup$
– Servaes
Mar 29 at 0:39
add a comment |
$begingroup$
There is no need to involve probabilities, just a bit of analytic number theory suffices. I don't know how familiar you are with $L$-functions, so I'll leave some details open for now. If any part needs clarification, let me know.
For every positive integer $n$ define
$$f(n):=sum_a,b=1^nfrac1gcd(a,b)
qquadtext and qquad
g(n):=sum_dmid nfracvarphi(tfracnd)d$$ so that for all $n>1$ we have
begineqnarray*
f(n)-f(n-1)
&=&-frac1n+2sum_c=1^nfrac1gcd(c,n)\
&=&-frac1n+2sum_dmid nfracvarphi(tfracnd)d\
&=&-frac1n+2g(n).
endeqnarray*
Because $lim_ntoinftyfrac1n^2sum_k=1^n-frac1k=0$, this already shows that
$$lim_ntoinftyfrac1n^2f(n)=2lim_ntoinftyfrac1n^2sum_k=1^ng(n).$$
Note that $g$ is a multiplicative function, and in fact $g=iotaastvarphi$ where $iota(n):=tfrac1n$ and $ast$ denotes the Dirichlet convolution of multiplicative functions. This immediately shows that the $L$-function associated to $g$ converges for all $sinBbbC$ with $operatornameRe(s)>2$ and that for all such $sinBbbC$ we have
$$L_g(s)=L_iota(s)L_varphi(s)=zeta(s+1)L_varphi(s).$$
It is easy to see that $L_iota(s)=zeta(s+1)$ for these $s$, and it is a basic exercise to show that
$$L_varphi(s)=fraczeta(s-1)zeta(s),$$
whenever $operatornameRe(s)>2$. Then by a Tauberian theorem, for example Wiener-Ikehara, we have
begineqnarray*
lim_ntoinftyfrac1n^2f(n)
&=&2operatornameRes(L_g,2)\
&=&2cdotoperatornameRes(zeta(s+1),2)cdotoperatornameRes(L_varphi,2)\
&=&2cdotzeta(3)cdotfrac12zeta(2)\
&=&fraczeta(3)zeta(2).
endeqnarray*
answered Mar 19 at 21:00
ServaesServaes
29.9k342101
$endgroup$
$begingroup$
Not familiar with $L$-functions at all. Anyway, thanks for your answer.
$endgroup$
– Crostul
Mar 19 at 23:25
$begingroup$
the point is that $L_g(s) = zeta(s+1)zeta(s-1)/zeta(s)$ then $L_g(s) = sum_nge 1 g(n) n^-s=sint_1^infty sum_n le x g(n) x^-s-1dx$ implies $sum_n le x g(n) = frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ and the Tauberian theorem is about using analyticity and boundedness to shift $int_3-iinfty^3+iinfty$ to the left, same method as the one developed for the PNT @Crostul
$endgroup$
– reuns
Mar 29 at 0:31
$begingroup$
@reuns I have included a mention of this, I won't provide a proof here.
$endgroup$
– Servaes
Mar 29 at 0:35
$begingroup$
The main difficulty is that $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ doesn't converge absolutely, we can look instead at $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ss^2ds$ or $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^s+1s(s+1)ds$ then deduce the asymptotic for the original one
$endgroup$
– reuns
Mar 29 at 0:37
$begingroup$
@reuns I agree that there are some subtle details in applying the theorem, but seeing that the OP is not familiar with $L$-functions at all, I thought I'd omit those until asked for.
$endgroup$
– Servaes
Mar 29 at 0:39
add a comment |
$begingroup$
There is no need to involve probabilities, just a bit of analytic number theory suffices. I don't know how familiar you are with $L$-functions, so I'll leave some details open for now. If any part needs clarification, let me know.
For every positive integer $n$ define
$$f(n):=sum_a,b=1^nfrac1gcd(a,b)
qquadtext and qquad
g(n):=sum_dmid nfracvarphi(tfracnd)d$$ so that for all $n>1$ we have
begineqnarray*
f(n)-f(n-1)
&=&-frac1n+2sum_c=1^nfrac1gcd(c,n)\
&=&-frac1n+2sum_dmid nfracvarphi(tfracnd)d\
&=&-frac1n+2g(n).
endeqnarray*
Because $lim_ntoinftyfrac1n^2sum_k=1^n-frac1k=0$, this already shows that
$$lim_ntoinftyfrac1n^2f(n)=2lim_ntoinftyfrac1n^2sum_k=1^ng(n).$$
Note that $g$ is a multiplicative function, and in fact $g=iotaastvarphi$ where $iota(n):=tfrac1n$ and $ast$ denotes the Dirichlet convolution of multiplicative functions. This immediately shows that the $L$-function associated to $g$ converges for all $sinBbbC$ with $operatornameRe(s)>2$ and that for all such $sinBbbC$ we have
$$L_g(s)=L_iota(s)L_varphi(s)=zeta(s+1)L_varphi(s).$$
It is easy to see that $L_iota(s)=zeta(s+1)$ for these $s$, and it is a basic exercise to show that
$$L_varphi(s)=fraczeta(s-1)zeta(s),$$
whenever $operatornameRe(s)>2$. Then by a Tauberian theorem, for example Wiener-Ikehara, we have
begineqnarray*
lim_ntoinftyfrac1n^2f(n)
&=&2operatornameRes(L_g,2)\
&=&2cdotoperatornameRes(zeta(s+1),2)cdotoperatornameRes(L_varphi,2)\
&=&2cdotzeta(3)cdotfrac12zeta(2)\
&=&fraczeta(3)zeta(2).
endeqnarray*
answered Mar 19 at 21:00
ServaesServaes
29.9k342101
$endgroup$
There is no need to involve probabilities, just a bit of analytic number theory suffices. I don't know how familiar you are with $L$-functions, so I'll leave some details open for now. If any part needs clarification, let me know.
For every positive integer $n$ define
$$f(n):=sum_a,b=1^nfrac1gcd(a,b)
qquadtext and qquad
g(n):=sum_dmid nfracvarphi(tfracnd)d$$ so that for all $n>1$ we have
begineqnarray*
f(n)-f(n-1)
&=&-frac1n+2sum_c=1^nfrac1gcd(c,n)\
&=&-frac1n+2sum_dmid nfracvarphi(tfracnd)d\
&=&-frac1n+2g(n).
endeqnarray*
Because $lim_ntoinftyfrac1n^2sum_k=1^n-frac1k=0$, this already shows that
$$lim_ntoinftyfrac1n^2f(n)=2lim_ntoinftyfrac1n^2sum_k=1^ng(n).$$
Note that $g$ is a multiplicative function, and in fact $g=iotaastvarphi$ where $iota(n):=tfrac1n$ and $ast$ denotes the Dirichlet convolution of multiplicative functions. This immediately shows that the $L$-function associated to $g$ converges for all $sinBbbC$ with $operatornameRe(s)>2$ and that for all such $sinBbbC$ we have
$$L_g(s)=L_iota(s)L_varphi(s)=zeta(s+1)L_varphi(s).$$
It is easy to see that $L_iota(s)=zeta(s+1)$ for these $s$, and it is a basic exercise to show that
$$L_varphi(s)=fraczeta(s-1)zeta(s),$$
whenever $operatornameRe(s)>2$. Then by a Tauberian theorem, for example Wiener-Ikehara, we have
begineqnarray*
lim_ntoinftyfrac1n^2f(n)
&=&2operatornameRes(L_g,2)\
&=&2cdotoperatornameRes(zeta(s+1),2)cdotoperatornameRes(L_varphi,2)\
&=&2cdotzeta(3)cdotfrac12zeta(2)\
&=&fraczeta(3)zeta(2).
endeqnarray*
answered Mar 19 at 21:00
ServaesServaes
29.9k342101
edited Mar 29 at 0:35
answered Mar 19 at 21:00
ServaesServaes
29.9k342101
answered Mar 19 at 21:00
ServaesServaes
29.9k342101
answered Mar 19 at 21:00
ServaesServaes
29.9k342101
29.9k342101
$begingroup$
Not familiar with $L$-functions at all. Anyway, thanks for your answer.
$endgroup$
– Crostul
Mar 19 at 23:25
$begingroup$
the point is that $L_g(s) = zeta(s+1)zeta(s-1)/zeta(s)$ then $L_g(s) = sum_nge 1 g(n) n^-s=sint_1^infty sum_n le x g(n) x^-s-1dx$ implies $sum_n le x g(n) = frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ and the Tauberian theorem is about using analyticity and boundedness to shift $int_3-iinfty^3+iinfty$ to the left, same method as the one developed for the PNT @Crostul
$endgroup$
– reuns
Mar 29 at 0:31
$begingroup$
@reuns I have included a mention of this, I won't provide a proof here.
$endgroup$
– Servaes
Mar 29 at 0:35
$begingroup$
The main difficulty is that $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ doesn't converge absolutely, we can look instead at $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ss^2ds$ or $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^s+1s(s+1)ds$ then deduce the asymptotic for the original one
$endgroup$
– reuns
Mar 29 at 0:37
$begingroup$
@reuns I agree that there are some subtle details in applying the theorem, but seeing that the OP is not familiar with $L$-functions at all, I thought I'd omit those until asked for.
$endgroup$
– Servaes
Mar 29 at 0:39
add a comment |
$begingroup$
Not familiar with $L$-functions at all. Anyway, thanks for your answer.
$endgroup$
– Crostul
Mar 19 at 23:25
$begingroup$
the point is that $L_g(s) = zeta(s+1)zeta(s-1)/zeta(s)$ then $L_g(s) = sum_nge 1 g(n) n^-s=sint_1^infty sum_n le x g(n) x^-s-1dx$ implies $sum_n le x g(n) = frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ and the Tauberian theorem is about using analyticity and boundedness to shift $int_3-iinfty^3+iinfty$ to the left, same method as the one developed for the PNT @Crostul
$endgroup$
– reuns
Mar 29 at 0:31
$begingroup$
@reuns I have included a mention of this, I won't provide a proof here.
$endgroup$
– Servaes
Mar 29 at 0:35
$begingroup$
The main difficulty is that $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ doesn't converge absolutely, we can look instead at $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ss^2ds$ or $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^s+1s(s+1)ds$ then deduce the asymptotic for the original one
$endgroup$
– reuns
Mar 29 at 0:37
$begingroup$
@reuns I agree that there are some subtle details in applying the theorem, but seeing that the OP is not familiar with $L$-functions at all, I thought I'd omit those until asked for.
$endgroup$
– Servaes
Mar 29 at 0:39
$begingroup$
Not familiar with $L$-functions at all. Anyway, thanks for your answer.
$endgroup$
– Crostul
Mar 19 at 23:25
$begingroup$
Not familiar with $L$-functions at all. Anyway, thanks for your answer.
$endgroup$
– Crostul
Mar 19 at 23:25
$begingroup$
the point is that $L_g(s) = zeta(s+1)zeta(s-1)/zeta(s)$ then $L_g(s) = sum_nge 1 g(n) n^-s=sint_1^infty sum_n le x g(n) x^-s-1dx$ implies $sum_n le x g(n) = frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ and the Tauberian theorem is about using analyticity and boundedness to shift $int_3-iinfty^3+iinfty$ to the left, same method as the one developed for the PNT @Crostul
$endgroup$
– reuns
Mar 29 at 0:31
$begingroup$
the point is that $L_g(s) = zeta(s+1)zeta(s-1)/zeta(s)$ then $L_g(s) = sum_nge 1 g(n) n^-s=sint_1^infty sum_n le x g(n) x^-s-1dx$ implies $sum_n le x g(n) = frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ and the Tauberian theorem is about using analyticity and boundedness to shift $int_3-iinfty^3+iinfty$ to the left, same method as the one developed for the PNT @Crostul
$endgroup$
– reuns
Mar 29 at 0:31
$begingroup$
@reuns I have included a mention of this, I won't provide a proof here.
$endgroup$
– Servaes
Mar 29 at 0:35
$begingroup$
@reuns I have included a mention of this, I won't provide a proof here.
$endgroup$
– Servaes
Mar 29 at 0:35
$begingroup$
The main difficulty is that $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ doesn't converge absolutely, we can look instead at $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ss^2ds$ or $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^s+1s(s+1)ds$ then deduce the asymptotic for the original one
$endgroup$
– reuns
Mar 29 at 0:37
$begingroup$
The main difficulty is that $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ssds$ doesn't converge absolutely, we can look instead at $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^ss^2ds$ or $frac12ipi int_3-iinfty^3+iinfty L_g(s) fracx^s+1s(s+1)ds$ then deduce the asymptotic for the original one
$endgroup$
– reuns
Mar 29 at 0:37
$begingroup$
@reuns I agree that there are some subtle details in applying the theorem, but seeing that the OP is not familiar with $L$-functions at all, I thought I'd omit those until asked for.
$endgroup$
– Servaes
Mar 29 at 0:39
$begingroup$
@reuns I agree that there are some subtle details in applying the theorem, but seeing that the OP is not familiar with $L$-functions at all, I thought I'd omit those until asked for.
$endgroup$
– Servaes
Mar 29 at 0:39
add a comment |
$begingroup$
Your doubts are justified: $sum_d=1^n fraczeta(2)^-1d^2$ is just a sum of limiting probabilities, so it should (and does) yield $1$ in the limit. The sum you want to compute is of values of $frac1gcd(a,b)$, so you need to weight the possible values of that expression by their probabilities. This gives
$$
sum_d=1^n fraczeta(2)^-1d^2cdotfrac1dtofraczeta(3)zeta(2)approx 0.730763
$$
A numerical experimentat (with $n=1000$) gives a result close to $0.731$, so this seems likely to be correct.
answered Mar 17 at 11:18
FredHFredH
3,6851023
$endgroup$
$begingroup$
Thank you for your answer! Indeed I was having doubts while writing my question. Anyway, this is just heuristics, now what I'm looking for is a proof.
$endgroup$
– Crostul
Mar 17 at 11:23
add a comment |
$begingroup$
Your doubts are justified: $sum_d=1^n fraczeta(2)^-1d^2$ is just a sum of limiting probabilities, so it should (and does) yield $1$ in the limit. The sum you want to compute is of values of $frac1gcd(a,b)$, so you need to weight the possible values of that expression by their probabilities. This gives
$$
sum_d=1^n fraczeta(2)^-1d^2cdotfrac1dtofraczeta(3)zeta(2)approx 0.730763
$$
A numerical experimentat (with $n=1000$) gives a result close to $0.731$, so this seems likely to be correct.
answered Mar 17 at 11:18
FredHFredH
3,6851023
$endgroup$
$begingroup$
Thank you for your answer! Indeed I was having doubts while writing my question. Anyway, this is just heuristics, now what I'm looking for is a proof.
$endgroup$
– Crostul
Mar 17 at 11:23
add a comment |
$begingroup$
Your doubts are justified: $sum_d=1^n fraczeta(2)^-1d^2$ is just a sum of limiting probabilities, so it should (and does) yield $1$ in the limit. The sum you want to compute is of values of $frac1gcd(a,b)$, so you need to weight the possible values of that expression by their probabilities. This gives
$$
sum_d=1^n fraczeta(2)^-1d^2cdotfrac1dtofraczeta(3)zeta(2)approx 0.730763
$$
A numerical experimentat (with $n=1000$) gives a result close to $0.731$, so this seems likely to be correct.
answered Mar 17 at 11:18
FredHFredH
3,6851023
$endgroup$
Your doubts are justified: $sum_d=1^n fraczeta(2)^-1d^2$ is just a sum of limiting probabilities, so it should (and does) yield $1$ in the limit. The sum you want to compute is of values of $frac1gcd(a,b)$, so you need to weight the possible values of that expression by their probabilities. This gives
$$
sum_d=1^n fraczeta(2)^-1d^2cdotfrac1dtofraczeta(3)zeta(2)approx 0.730763
$$
A numerical experimentat (with $n=1000$) gives a result close to $0.731$, so this seems likely to be correct.
answered Mar 17 at 11:18
FredHFredH
3,6851023
answered Mar 17 at 11:18
FredHFredH
3,6851023
answered Mar 17 at 11:18
FredHFredH
3,6851023
answered Mar 17 at 11:18
FredHFredH
3,6851023
3,6851023
$begingroup$
Thank you for your answer! Indeed I was having doubts while writing my question. Anyway, this is just heuristics, now what I'm looking for is a proof.
$endgroup$
– Crostul
Mar 17 at 11:23
add a comment |
$begingroup$
Thank you for your answer! Indeed I was having doubts while writing my question. Anyway, this is just heuristics, now what I'm looking for is a proof.
$endgroup$
– Crostul
Mar 17 at 11:23
$begingroup$
Thank you for your answer! Indeed I was having doubts while writing my question. Anyway, this is just heuristics, now what I'm looking for is a proof.
$endgroup$
– Crostul
Mar 17 at 11:23
$begingroup$
Thank you for your answer! Indeed I was having doubts while writing my question. Anyway, this is just heuristics, now what I'm looking for is a proof.
$endgroup$
– Crostul
Mar 17 at 11:23
add a comment |
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$begingroup$
The limit cannot be $1$ since for each $n$, there are at least $frac(n-1)^24$ pairs of even numbers $a,b$ with $textgcd(a,b)ge 2$. So the limit is at most $1cdot frac 3 4+frac 12 frac 14 =frac 7 8.$
$endgroup$
– Song
Mar 17 at 11:32