Why shouldn't this prove the Prime Number Theorem? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?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?Any way to prove Prime Number Theorem using Hyperbolic Geometry?Any ways to Simplify Daboussi's Argument for Prime Number Theorem?Effective prime number theoremIs the number $sum_ptext primep^-2$ known to be irrational?Landau's theorem using nth roots
Why shouldn't this prove the Prime Number Theorem?
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?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?Any way to prove Prime Number Theorem using Hyperbolic Geometry?Any ways to Simplify Daboussi's Argument for Prime Number Theorem?Effective prime number theoremIs the number $sum_ptext primep^-2$ known to be irrational?Landau'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
asked 5 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 $sum_n=1^infty fracmu(n)n^k$ 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
asked 5 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 $sum_n leq x fracmu(n)n = o(1)$. 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
5 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
5 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
4 hours ago
1
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
4 hours ago
add a comment |
$begingroup$
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ 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
asked 5 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 $sum_n=1^infty fracmu(n)n^k$ 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
nt.number-theory prime-numbers
asked 5 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 5 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 5 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 5 hours ago
Fourton.Fourton.
422
asked 5 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 $sum_n leq x fracmu(n)n = o(1)$. 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
5 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
5 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
4 hours ago
1
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
4 hours ago
add a comment |
11
$begingroup$
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
5 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
5 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
4 hours ago
1
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
4 hours ago
11
11
$begingroup$
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
5 hours ago
$begingroup$
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
5 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
5 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
5 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
4 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
4 hours ago
1
1
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
4 hours ago
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
4 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 $sum_n=1^infty fracmu(n)n^k$ 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
$$sum_n=1^infty fracmu(n)n=o(1),$$
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.
$endgroup$
add a comment |
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1 Answer
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$begingroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ 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
$$sum_n=1^infty fracmu(n)n=o(1),$$
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.
$endgroup$
add a comment |
$begingroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ 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
$$sum_n=1^infty fracmu(n)n=o(1),$$
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.
$endgroup$
add a comment |
$begingroup$
You ask:
Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ 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
$$sum_n=1^infty fracmu(n)n=o(1),$$
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.
$endgroup$
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
$$sum_n=1^infty fracmu(n)n=o(1),$$
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.
answered 5 mins ago
community wiki
kodlu
add a comment |
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.
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 $sum_n leq x fracmu(n)n = o(1)$. 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
5 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
5 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
4 hours ago
1
$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
4 hours ago