Why shouldn't this prove the Prime Number Theorem?Heuristic argument for the prime number theorem?Why is the Chebyshev function relevant to the Prime Number TheoremWhy could Mertens not prove the prime number theorem?Probability that randomly chosen integers from a restricted set of natural numbers are coprimeCan the following quantitative version of Chen's theorem be obtained?The shortest interval for which the prime number theorem holdsAny way to prove Prime Number Theorem using Hyperbolic Geometry?Any ways to Simplify Daboussi's Argument for Prime Number Theorem?Effective prime number theoremLandau's theorem using nth roots
Why shouldn't this prove the Prime Number Theorem?
Heuristic argument for the prime number theorem?Why is the Chebyshev function relevant to the Prime Number TheoremWhy could Mertens not prove the prime number theorem?Probability that randomly chosen integers from a restricted set of natural numbers are coprimeCan the following quantitative version of Chen's theorem be obtained?The shortest interval for which the prime number theorem holdsAny way to prove Prime Number Theorem using Hyperbolic Geometry?Any ways to Simplify Daboussi's Argument for Prime Number Theorem?Effective prime number theoremLandau's theorem using nth roots
$begingroup$
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ can be interpreted as the probability that a randomly chosen integer is $k$-free.
Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form
$$sum_n=1^infty fracmu(n)n=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
nt.number-theory prime-numbers prime-number-theorem
edited 2 days ago
Martin Sleziak
3,14032231
asked Apr 20 at 21:45
Fourton.Fourton.
814
New contributor
Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
$begingroup$
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ can be interpreted as the probability that a randomly chosen integer is $k$-free.
Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form
$$sum_n=1^infty fracmu(n)n=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
nt.number-theory prime-numbers prime-number-theorem
edited 2 days ago
Martin Sleziak
3,14032231
asked Apr 20 at 21:45
Fourton.Fourton.
814
New contributor
Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
19
$begingroup$
It is true that the PNT is equivalent to $sum_n leq x fracmu(n)n = o(1)$. It is also relatively easy to prove that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = 0$. The hard part is proving that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = lim_x to infty sum_n leq x fracmu(n)n$. This is highly nontrivial!
$endgroup$
– Peter Humphries
Apr 20 at 21:48
14
$begingroup$
In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
$endgroup$
– Wojowu
Apr 20 at 21:48
10
$begingroup$
I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
$endgroup$
– Gro-Tsen
Apr 20 at 22:41
2
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
Apr 20 at 22:43
$begingroup$
@YemonChoi, is that a new nickname for @Gro-Tsen? :-)
$endgroup$
– LSpice
yesterday
|
$begingroup$
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ can be interpreted as the probability that a randomly chosen integer is $k$-free.
Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form
$$sum_n=1^infty fracmu(n)n=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
nt.number-theory prime-numbers prime-number-theorem
edited 2 days ago
Martin Sleziak
3,14032231
asked Apr 20 at 21:45
Fourton.Fourton.
814
New contributor
Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ can be interpreted as the probability that a randomly chosen integer is $k$-free.
Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form
$$sum_n=1^infty fracmu(n)n=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
nt.number-theory prime-numbers prime-number-theorem
nt.number-theory prime-numbers prime-number-theorem
edited 2 days ago
Martin Sleziak
3,14032231
asked Apr 20 at 21:45
Fourton.Fourton.
814
New contributor
Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Martin Sleziak
3,14032231
asked Apr 20 at 21:45
Fourton.Fourton.
814
New contributor
Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Martin Sleziak
3,14032231
edited 2 days ago
Martin Sleziak
3,14032231
edited 2 days ago
Martin Sleziak
3,14032231
3,14032231
asked Apr 20 at 21:45
Fourton.Fourton.
814
New contributor
Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 20 at 21:45
Fourton.Fourton.
814
asked Apr 20 at 21:45
Fourton.Fourton.
814
814
New contributor
Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
19
$begingroup$
It is true that the PNT is equivalent to $sum_n leq x fracmu(n)n = o(1)$. It is also relatively easy to prove that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = 0$. The hard part is proving that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = lim_x to infty sum_n leq x fracmu(n)n$. This is highly nontrivial!
$endgroup$
– Peter Humphries
Apr 20 at 21:48
14
$begingroup$
In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
$endgroup$
– Wojowu
Apr 20 at 21:48
10
$begingroup$
I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
$endgroup$
– Gro-Tsen
Apr 20 at 22:41
2
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
Apr 20 at 22:43
$begingroup$
@YemonChoi, is that a new nickname for @Gro-Tsen? :-)
$endgroup$
– LSpice
yesterday
|
19
$begingroup$
It is true that the PNT is equivalent to $sum_n leq x fracmu(n)n = o(1)$. It is also relatively easy to prove that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = 0$. The hard part is proving that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = lim_x to infty sum_n leq x fracmu(n)n$. This is highly nontrivial!
$endgroup$
– Peter Humphries
Apr 20 at 21:48
14
$begingroup$
In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
$endgroup$
– Wojowu
Apr 20 at 21:48
10
$begingroup$
I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
$endgroup$
– Gro-Tsen
Apr 20 at 22:41
2
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
Apr 20 at 22:43
$begingroup$
@YemonChoi, is that a new nickname for @Gro-Tsen? :-)
$endgroup$
– LSpice
yesterday
19
19
$begingroup$
It is true that the PNT is equivalent to $sum_n leq x fracmu(n)n = o(1)$. It is also relatively easy to prove that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = 0$. The hard part is proving that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = lim_x to infty sum_n leq x fracmu(n)n$. This is highly nontrivial!
$endgroup$
– Peter Humphries
Apr 20 at 21:48
$begingroup$
It is true that the PNT is equivalent to $sum_n leq x fracmu(n)n = o(1)$. It is also relatively easy to prove that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = 0$. The hard part is proving that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = lim_x to infty sum_n leq x fracmu(n)n$. This is highly nontrivial!
$endgroup$
– Peter Humphries
Apr 20 at 21:48
14
14
$begingroup$
In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
$endgroup$
– Wojowu
Apr 20 at 21:48
$begingroup$
In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
$endgroup$
– Wojowu
Apr 20 at 21:48
10
10
$begingroup$
I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
$endgroup$
– Gro-Tsen
Apr 20 at 22:41
$begingroup$
I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
$endgroup$
– Gro-Tsen
Apr 20 at 22:41
2
2
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
Apr 20 at 22:43
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
Apr 20 at 22:43
$begingroup$
@YemonChoi, is that a new nickname for @Gro-Tsen? :-)
$endgroup$
– LSpice
yesterday
$begingroup$
@YemonChoi, is that a new nickname for @Gro-Tsen? :-)
$endgroup$
– LSpice
yesterday
|
1 Answer
1
active
oldest
votes
$begingroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ can be interpreted as the probability that a randomly chosen integer is $k$-free.
Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form
$$sum_n=1^infty fracmu(n)n=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to
$$lim_x to infty sum_nleq x fracmu(n)n=0,$$
and it is relatively easy to prove that
$$lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s=0.$$
The real difficulty lies in proving that
$$lim_xrightarrow infty sum_nleq x fracmu(n)n=
lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s,$$
which is highly nontrivial and requires intricate arguments.
In particular, as pointed out by @TerryTao in the comments:
if $tneq 0$ is real, then
$$ lim_srightarrow 1^+ sum_n=1^infty fracn^itn^s,$$
can be shown to converge to a finite value, whereas
$$lim_xrightarrow infty sum_nleq x fracn^itn$$
is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the PNT.
$endgroup$
13
$begingroup$
Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
$endgroup$
– Nate Eldredge
2 days ago
4
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
2 days ago
13
$begingroup$
In particular, if $t$ a non-zero real, then $lim_s to 1^+ sum_n=1^infty fracn^itn^s$ can be shown to converge to a finite value, whereas $lim_x to infty sum_n leq x fracn^itn$ is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the prime number theorem.
$endgroup$
– Terry Tao
2 days ago
|
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ can be interpreted as the probability that a randomly chosen integer is $k$-free.
Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form
$$sum_n=1^infty fracmu(n)n=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to
$$lim_x to infty sum_nleq x fracmu(n)n=0,$$
and it is relatively easy to prove that
$$lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s=0.$$
The real difficulty lies in proving that
$$lim_xrightarrow infty sum_nleq x fracmu(n)n=
lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s,$$
which is highly nontrivial and requires intricate arguments.
In particular, as pointed out by @TerryTao in the comments:
if $tneq 0$ is real, then
$$ lim_srightarrow 1^+ sum_n=1^infty fracn^itn^s,$$
can be shown to converge to a finite value, whereas
$$lim_xrightarrow infty sum_nleq x fracn^itn$$
is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the PNT.
$endgroup$
13
$begingroup$
Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
$endgroup$
– Nate Eldredge
2 days ago
4
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
2 days ago
13
$begingroup$
In particular, if $t$ a non-zero real, then $lim_s to 1^+ sum_n=1^infty fracn^itn^s$ can be shown to converge to a finite value, whereas $lim_x to infty sum_n leq x fracn^itn$ is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the prime number theorem.
$endgroup$
– Terry Tao
2 days ago
|
$begingroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ can be interpreted as the probability that a randomly chosen integer is $k$-free.
Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form
$$sum_n=1^infty fracmu(n)n=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to
$$lim_x to infty sum_nleq x fracmu(n)n=0,$$
and it is relatively easy to prove that
$$lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s=0.$$
The real difficulty lies in proving that
$$lim_xrightarrow infty sum_nleq x fracmu(n)n=
lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s,$$
which is highly nontrivial and requires intricate arguments.
In particular, as pointed out by @TerryTao in the comments:
if $tneq 0$ is real, then
$$ lim_srightarrow 1^+ sum_n=1^infty fracn^itn^s,$$
can be shown to converge to a finite value, whereas
$$lim_xrightarrow infty sum_nleq x fracn^itn$$
is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the PNT.
$endgroup$
13
$begingroup$
Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
$endgroup$
– Nate Eldredge
2 days ago
4
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
2 days ago
13
$begingroup$
In particular, if $t$ a non-zero real, then $lim_s to 1^+ sum_n=1^infty fracn^itn^s$ can be shown to converge to a finite value, whereas $lim_x to infty sum_n leq x fracn^itn$ is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the prime number theorem.
$endgroup$
– Terry Tao
2 days ago
|
$begingroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ can be interpreted as the probability that a randomly chosen integer is $k$-free.
Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form
$$sum_n=1^infty fracmu(n)n=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to
$$lim_x to infty sum_nleq x fracmu(n)n=0,$$
and it is relatively easy to prove that
$$lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s=0.$$
The real difficulty lies in proving that
$$lim_xrightarrow infty sum_nleq x fracmu(n)n=
lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s,$$
which is highly nontrivial and requires intricate arguments.
In particular, as pointed out by @TerryTao in the comments:
if $tneq 0$ is real, then
$$ lim_srightarrow 1^+ sum_n=1^infty fracn^itn^s,$$
can be shown to converge to a finite value, whereas
$$lim_xrightarrow infty sum_nleq x fracn^itn$$
is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the PNT.
$endgroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ can be interpreted as the probability that a randomly chosen integer is $k$-free.
Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form
$$sum_n=1^infty fracmu(n)n=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to
$$lim_x to infty sum_nleq x fracmu(n)n=0,$$
and it is relatively easy to prove that
$$lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s=0.$$
The real difficulty lies in proving that
$$lim_xrightarrow infty sum_nleq x fracmu(n)n=
lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s,$$
which is highly nontrivial and requires intricate arguments.
In particular, as pointed out by @TerryTao in the comments:
if $tneq 0$ is real, then
$$ lim_srightarrow 1^+ sum_n=1^infty fracn^itn^s,$$
can be shown to converge to a finite value, whereas
$$lim_xrightarrow infty sum_nleq x fracn^itn$$
is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the PNT.
edited 2 days ago
community wiki
3 revs, 2 users 98%
kodlu
13
$begingroup$
Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
$endgroup$
– Nate Eldredge
2 days ago
4
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
2 days ago
13
$begingroup$
In particular, if $t$ a non-zero real, then $lim_s to 1^+ sum_n=1^infty fracn^itn^s$ can be shown to converge to a finite value, whereas $lim_x to infty sum_n leq x fracn^itn$ is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the prime number theorem.
$endgroup$
– Terry Tao
2 days ago
|
13
$begingroup$
Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
$endgroup$
– Nate Eldredge
2 days ago
4
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
2 days ago
13
$begingroup$
In particular, if $t$ a non-zero real, then $lim_s to 1^+ sum_n=1^infty fracn^itn^s$ can be shown to converge to a finite value, whereas $lim_x to infty sum_n leq x fracn^itn$ is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the prime number theorem.
$endgroup$
– Terry Tao
2 days ago
13
13
$begingroup$
Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
$endgroup$
– Nate Eldredge
2 days ago
$begingroup$
Analyst's life story: you have two limiting operations (limit, infinite sum, integral, derivative, etc), and if only you could interchange them, you'd have your result; but in order to justify doing so, you need some hard estimates...
$endgroup$
– Nate Eldredge
2 days ago
4
4
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
2 days ago
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
2 days ago
13
13
$begingroup$
In particular, if $t$ a non-zero real, then $lim_s to 1^+ sum_n=1^infty fracn^itn^s$ can be shown to converge to a finite value, whereas $lim_x to infty sum_n leq x fracn^itn$ is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the prime number theorem.
$endgroup$
– Terry Tao
2 days ago
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In particular, if $t$ a non-zero real, then $lim_s to 1^+ sum_n=1^infty fracn^itn^s$ can be shown to converge to a finite value, whereas $lim_x to infty sum_n leq x fracn^itn$ is undefined. So at a bare minimum one has to somehow stop $mu(n)$ from "pretending" to be like $n^it$. This turns out to be basically equivalent to preventing $zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the prime number theorem.
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– Terry Tao
2 days ago
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It is true that the PNT is equivalent to $sum_n leq x fracmu(n)n = o(1)$. It is also relatively easy to prove that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = 0$. The hard part is proving that $lim_s searrow 1 sum_n = 1^infty fracmu(n)n^s = lim_x to infty sum_n leq x fracmu(n)n$. This is highly nontrivial!
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– Peter Humphries
Apr 20 at 21:48
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In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
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– Wojowu
Apr 20 at 21:48
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I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
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– Gro-Tsen
Apr 20 at 22:41
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I agree with Fourton and have voted accordingly
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– Yemon Choi
Apr 20 at 22:43
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@YemonChoi, is that a new nickname for @Gro-Tsen? :-)
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– LSpice
yesterday