Why shouldn't this prove the Prime Number Theorem?
$begingroup$
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ 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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
nt.number-theory prime-numbers prime-number-theorem
edited 20 mins ago
Martin Sleziak
3,13032231
asked 7 hours ago
Fourton.Fourton.
422
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.
$endgroup$
add a comment |
$begingroup$
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ 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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
nt.number-theory prime-numbers prime-number-theorem
edited 20 mins ago
Martin Sleziak
3,13032231
asked 7 hours ago
Fourton.Fourton.
422
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.
$endgroup$
11
$begingroup$
It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
$endgroup$
– Peter Humphries
7 hours ago
11
$begingroup$
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
7 hours ago
6
$begingroup$
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
6 hours ago
2
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
6 hours ago
add a comment |
$begingroup$
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ 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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
nt.number-theory prime-numbers prime-number-theorem
edited 20 mins ago
Martin Sleziak
3,13032231
asked 7 hours ago
Fourton.Fourton.
422
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 $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ 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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$
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 20 mins ago
Martin Sleziak
3,13032231
asked 7 hours ago
Fourton.Fourton.
422
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 20 mins ago
Martin Sleziak
3,13032231
asked 7 hours ago
Fourton.Fourton.
422
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 20 mins ago
Martin Sleziak
3,13032231
edited 20 mins ago
Martin Sleziak
3,13032231
edited 20 mins ago
Martin Sleziak
3,13032231
3,13032231
asked 7 hours ago
Fourton.Fourton.
422
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 7 hours ago
Fourton.Fourton.
422
asked 7 hours ago
Fourton.Fourton.
422
422
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.
11
$begingroup$
It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
$endgroup$
– Peter Humphries
7 hours ago
11
$begingroup$
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
7 hours ago
6
$begingroup$
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
6 hours ago
2
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
6 hours ago
add a comment |
11
$begingroup$
It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
$endgroup$
– Peter Humphries
7 hours ago
11
$begingroup$
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
7 hours ago
6
$begingroup$
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
6 hours ago
2
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
6 hours ago
11
11
$begingroup$
It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
$endgroup$
– Peter Humphries
7 hours ago
$begingroup$
It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
$endgroup$
– Peter Humphries
7 hours ago
11
11
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
6
6
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
2
2
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
6 hours ago
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
6 hours ago
add a comment |
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 $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ 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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$
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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
and it is relatively easy to prove that
$$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
The real difficulty lies in proving that
$$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.
$endgroup$
3
$begingroup$
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 hours ago
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
1 hour ago
add a comment |
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$begingroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ 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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$
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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
and it is relatively easy to prove that
$$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
The real difficulty lies in proving that
$$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.
$endgroup$
3
$begingroup$
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 hours ago
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
1 hour ago
add a comment |
$begingroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ 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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$
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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
and it is relatively easy to prove that
$$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
The real difficulty lies in proving that
$$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.
$endgroup$
3
$begingroup$
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 hours ago
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
1 hour ago
add a comment |
$begingroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ 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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$
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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
and it is relatively easy to prove that
$$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
The real difficulty lies in proving that
$$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.
$endgroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ 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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$
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
$$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
and it is relatively easy to prove that
$$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
The real difficulty lies in proving that
$$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.
answered 2 hours ago
community wiki
kodlu
3
$begingroup$
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 hours ago
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
1 hour ago
add a comment |
3
$begingroup$
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 hours ago
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
1 hour ago
3
3
$begingroup$
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 hours 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 hours ago
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
1 hour ago
$begingroup$
@NateEldredge Or magic guarantees like (weak) compactness...
$endgroup$
– Yemon Choi
1 hour ago
add a comment |
Fourton. is a new contributor. Be nice, and check out our Code of Conduct.
Fourton. is a new contributor. Be nice, and check out our Code of Conduct.
Fourton. is a new contributor. Be nice, and check out our Code of Conduct.
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11
$begingroup$
It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
$endgroup$
– Peter Humphries
7 hours ago
11
$begingroup$
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
7 hours ago
6
$begingroup$
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
6 hours ago
2
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
6 hours ago