Are there any series whose convergence is unknown?
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Are there any infinite series about which we don't know whether it converges or not? Or are the convergence tests exhaustive, so that in the hands of a competent mathematician any series will eventually be shown to converge or diverge?
EDIT: People were kind enough to point out that ,without imposing restrictions on the terms, it's trivial to find such "open problem" sequences. So, to clarify, what I had in mind were sequences whose terms are composed of "simple" functions, the kind you would find in an introductory calculus text: exponential, factorial, etc.
real-analysis sequences-and-series soft-question infinity
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|
show 1 more comment
$begingroup$
Are there any infinite series about which we don't know whether it converges or not? Or are the convergence tests exhaustive, so that in the hands of a competent mathematician any series will eventually be shown to converge or diverge?
EDIT: People were kind enough to point out that ,without imposing restrictions on the terms, it's trivial to find such "open problem" sequences. So, to clarify, what I had in mind were sequences whose terms are composed of "simple" functions, the kind you would find in an introductory calculus text: exponential, factorial, etc.
real-analysis sequences-and-series soft-question infinity
$endgroup$
35
$begingroup$
Let $a_n = 1$ if $p_n+2$ is a prime where $p_n$ is the $n$th prime. The convergence of $sum_n a_n$ is equivalent to the twin prime conjecture. Of course, this isn't the likely sense in which the OP asked the question but maybe the question needs to be more specific?
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– Dinesh
Feb 5 '11 at 17:36
12
$begingroup$
The problem with this question is that one can encode a great deal of other mathematics as the question of whether some artificial series converges, as Dinesh as shown. One has to somehow restrict to a natural series, and it's far from clear what this even means.
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– Qiaochu Yuan
Feb 5 '11 at 17:44
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@Dinesh: In fact any statement on proving something is countably infinite could be restated using an indicator function like what you have stated. Anyway good one.
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– user17762
Feb 5 '11 at 17:47
2
$begingroup$
Thanks guys, all good points. I clarified things a bit. (My question reminds me of the Ali G interview where he asks "Will a computer ever be able to work out what is 999999999... multiplied by 999999...?".)
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– garageàtrois
Feb 5 '11 at 21:55
1
$begingroup$
The answer to whether a computer can work out... is you are asking for $(10^m-1)*(10^n-1)=10^{(m+n)}-10^m-10^n+1$ and I think Wolfram Alpha can do that already.
$endgroup$
– Ross Millikan
Jul 2 '11 at 3:40
|
show 1 more comment
$begingroup$
Are there any infinite series about which we don't know whether it converges or not? Or are the convergence tests exhaustive, so that in the hands of a competent mathematician any series will eventually be shown to converge or diverge?
EDIT: People were kind enough to point out that ,without imposing restrictions on the terms, it's trivial to find such "open problem" sequences. So, to clarify, what I had in mind were sequences whose terms are composed of "simple" functions, the kind you would find in an introductory calculus text: exponential, factorial, etc.
real-analysis sequences-and-series soft-question infinity
$endgroup$
Are there any infinite series about which we don't know whether it converges or not? Or are the convergence tests exhaustive, so that in the hands of a competent mathematician any series will eventually be shown to converge or diverge?
EDIT: People were kind enough to point out that ,without imposing restrictions on the terms, it's trivial to find such "open problem" sequences. So, to clarify, what I had in mind were sequences whose terms are composed of "simple" functions, the kind you would find in an introductory calculus text: exponential, factorial, etc.
real-analysis sequences-and-series soft-question infinity
real-analysis sequences-and-series soft-question infinity
edited Feb 8 '18 at 3:03
garageàtrois
asked Feb 5 '11 at 17:30
garageàtroisgarageàtrois
753268
753268
35
$begingroup$
Let $a_n = 1$ if $p_n+2$ is a prime where $p_n$ is the $n$th prime. The convergence of $sum_n a_n$ is equivalent to the twin prime conjecture. Of course, this isn't the likely sense in which the OP asked the question but maybe the question needs to be more specific?
$endgroup$
– Dinesh
Feb 5 '11 at 17:36
12
$begingroup$
The problem with this question is that one can encode a great deal of other mathematics as the question of whether some artificial series converges, as Dinesh as shown. One has to somehow restrict to a natural series, and it's far from clear what this even means.
$endgroup$
– Qiaochu Yuan
Feb 5 '11 at 17:44
$begingroup$
@Dinesh: In fact any statement on proving something is countably infinite could be restated using an indicator function like what you have stated. Anyway good one.
$endgroup$
– user17762
Feb 5 '11 at 17:47
2
$begingroup$
Thanks guys, all good points. I clarified things a bit. (My question reminds me of the Ali G interview where he asks "Will a computer ever be able to work out what is 999999999... multiplied by 999999...?".)
$endgroup$
– garageàtrois
Feb 5 '11 at 21:55
1
$begingroup$
The answer to whether a computer can work out... is you are asking for $(10^m-1)*(10^n-1)=10^{(m+n)}-10^m-10^n+1$ and I think Wolfram Alpha can do that already.
$endgroup$
– Ross Millikan
Jul 2 '11 at 3:40
|
show 1 more comment
35
$begingroup$
Let $a_n = 1$ if $p_n+2$ is a prime where $p_n$ is the $n$th prime. The convergence of $sum_n a_n$ is equivalent to the twin prime conjecture. Of course, this isn't the likely sense in which the OP asked the question but maybe the question needs to be more specific?
$endgroup$
– Dinesh
Feb 5 '11 at 17:36
12
$begingroup$
The problem with this question is that one can encode a great deal of other mathematics as the question of whether some artificial series converges, as Dinesh as shown. One has to somehow restrict to a natural series, and it's far from clear what this even means.
$endgroup$
– Qiaochu Yuan
Feb 5 '11 at 17:44
$begingroup$
@Dinesh: In fact any statement on proving something is countably infinite could be restated using an indicator function like what you have stated. Anyway good one.
$endgroup$
– user17762
Feb 5 '11 at 17:47
2
$begingroup$
Thanks guys, all good points. I clarified things a bit. (My question reminds me of the Ali G interview where he asks "Will a computer ever be able to work out what is 999999999... multiplied by 999999...?".)
$endgroup$
– garageàtrois
Feb 5 '11 at 21:55
1
$begingroup$
The answer to whether a computer can work out... is you are asking for $(10^m-1)*(10^n-1)=10^{(m+n)}-10^m-10^n+1$ and I think Wolfram Alpha can do that already.
$endgroup$
– Ross Millikan
Jul 2 '11 at 3:40
35
35
$begingroup$
Let $a_n = 1$ if $p_n+2$ is a prime where $p_n$ is the $n$th prime. The convergence of $sum_n a_n$ is equivalent to the twin prime conjecture. Of course, this isn't the likely sense in which the OP asked the question but maybe the question needs to be more specific?
$endgroup$
– Dinesh
Feb 5 '11 at 17:36
$begingroup$
Let $a_n = 1$ if $p_n+2$ is a prime where $p_n$ is the $n$th prime. The convergence of $sum_n a_n$ is equivalent to the twin prime conjecture. Of course, this isn't the likely sense in which the OP asked the question but maybe the question needs to be more specific?
$endgroup$
– Dinesh
Feb 5 '11 at 17:36
12
12
$begingroup$
The problem with this question is that one can encode a great deal of other mathematics as the question of whether some artificial series converges, as Dinesh as shown. One has to somehow restrict to a natural series, and it's far from clear what this even means.
$endgroup$
– Qiaochu Yuan
Feb 5 '11 at 17:44
$begingroup$
The problem with this question is that one can encode a great deal of other mathematics as the question of whether some artificial series converges, as Dinesh as shown. One has to somehow restrict to a natural series, and it's far from clear what this even means.
$endgroup$
– Qiaochu Yuan
Feb 5 '11 at 17:44
$begingroup$
@Dinesh: In fact any statement on proving something is countably infinite could be restated using an indicator function like what you have stated. Anyway good one.
$endgroup$
– user17762
Feb 5 '11 at 17:47
$begingroup$
@Dinesh: In fact any statement on proving something is countably infinite could be restated using an indicator function like what you have stated. Anyway good one.
$endgroup$
– user17762
Feb 5 '11 at 17:47
2
2
$begingroup$
Thanks guys, all good points. I clarified things a bit. (My question reminds me of the Ali G interview where he asks "Will a computer ever be able to work out what is 999999999... multiplied by 999999...?".)
$endgroup$
– garageàtrois
Feb 5 '11 at 21:55
$begingroup$
Thanks guys, all good points. I clarified things a bit. (My question reminds me of the Ali G interview where he asks "Will a computer ever be able to work out what is 999999999... multiplied by 999999...?".)
$endgroup$
– garageàtrois
Feb 5 '11 at 21:55
1
1
$begingroup$
The answer to whether a computer can work out... is you are asking for $(10^m-1)*(10^n-1)=10^{(m+n)}-10^m-10^n+1$ and I think Wolfram Alpha can do that already.
$endgroup$
– Ross Millikan
Jul 2 '11 at 3:40
$begingroup$
The answer to whether a computer can work out... is you are asking for $(10^m-1)*(10^n-1)=10^{(m+n)}-10^m-10^n+1$ and I think Wolfram Alpha can do that already.
$endgroup$
– Ross Millikan
Jul 2 '11 at 3:40
|
show 1 more comment
4 Answers
4
active
oldest
votes
$begingroup$
It is unknown whether
$$
sum_{n=1}^inftyfrac{1}{n^3sin^2n}
$$
converges or not. The difficulty here is that convergence depends on the term $nsin n$ not being too small, which in turn depends on how well $pi$ can be approximated by rational numbers. It is possible that, if $pi$ can be approximated `too well' enough by rationals, then this will diverge. See this MathOverflow question for a discussion of this particular series.
Another even simpler example of a sequence (no summation) for which it is not known whether it converges or not is
$$
x_n=frac{1}{n^2sin n}.
$$
We would expect this to tend to zero, but the proof is beyond what is currently known. Suppose that there were only finitely many rational numbers $p/q$ with $vert p/q-pivertle q^{-3+epsilon}$ (for any $epsilon > 0$), then $x_n$ would tend to zero at rate $O(n^{-epsilon})$. If, on the other hand, there were infinitely many rationals satisfying $vert p/q-pivertle q^{-3-epsilon}$, then infinitely many $x_n$ would be of order at least $n^epsilon$, so it diverges. This can be expressed in terms of the irrationality measure of $pi$. The sequence $x_n$ converges to zero if the irrationality measure of $pi$ is less than 3, and diverges if it is geater than 3. Currently, the best known bound for the irrationality measure is that it is no more than about $7.6063$ (see the link to the mathworld page above). It is expected that the irrationality measure of $pi$ is 2 (it is known that all but a zero-measure set of real numbers have irrationality measure 2). Therefore, it is expected that $x_n$ tends to zero, but there is currently no proof of this.
$endgroup$
$begingroup$
Mentioned at MO mathoverflow.net/questions/100265/…
$endgroup$
– Alexander Chervov
Jun 22 '12 at 13:16
9
$begingroup$
I find it quite incredible that such a simple sequence ($x_n$) has unknown convergence.
$endgroup$
– Anon
Jan 6 '17 at 2:32
1
$begingroup$
Also mentioned on Wikipedia and named “Flint Hills series” both there and in the MO post you referenced.
$endgroup$
– MvG
Jan 6 '17 at 8:47
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I am blown away by this.
$endgroup$
– Randall
Nov 1 '17 at 15:04
add a comment |
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As kind of a joke answer, but technically correct, and motivated by Chandru's deleted reply,
$$sum_{n=0}^infty sin(2pi n!,x)$$
where $x$ is the Euler-Mascheroni constant, or almost any other number whose rationality has not been settled. (If $x$ is rational, the series converges. The implication does not go the other way.)
$endgroup$
1
$begingroup$
Out of curiosity, what makes this a "joke" answer?
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– AlanH
May 17 '13 at 1:58
2
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There is no analysis of decay rate involved.
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– André Nicolas
May 17 '13 at 19:20
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The comma before "or almost any" confused me for a second, better to remove it imho, without it one understands immediately your use of x instead of the usual gamma. (there is an Oxford comma, but not really for "or")
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– PatrickT
Feb 26 '18 at 6:50
add a comment |
$begingroup$
It is unknown whether the series:
$$sum_n frac{(-1)^n n}{p_n}$$
converges. Here, $p_n$ is the $n$-th prime number. This problem is posed in Guy's book on unsolved problems in number theory and I am pretty sure that it originated from Erdős.
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3
$begingroup$
Very nice. +1 for Erdős.
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– Bumblebee
Feb 19 '16 at 10:52
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Huh, well, if you accept the prime number theorem, then this converges, else, it is unknowable.
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– Simply Beautiful Art
Mar 9 '17 at 14:48
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@SimplyBeautifulArt the prime number theorem does not imply convergence of this series.
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– Mustafa Said
Mar 13 '17 at 2:38
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Really? By the alternating test, all I need to do is show that $frac n{p_n}to0$.
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– Simply Beautiful Art
Mar 13 '17 at 12:36
11
$begingroup$
@SimplyBeautifulArt in order to use the alternating series test you must show that the terms of the series are monotonically decreasing.
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– Mustafa Said
Mar 17 '17 at 21:30
add a comment |
$begingroup$
The Riemann hypothesis is that $sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^{3+epsilon} n}$ converges for any $epsilon > 0$ (see here for a discussion of that $epsilon$).
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3
$begingroup$
Here, $Lambda(n)$ is the logarithm of the prime number $p$ such that $n$ is a power of $p$, if any, or $0$ if $n$ is not a prime power. (It says so in the first link, but I thought it helpful to have it on this page.) I also presume that all of the logarithms are natural logarithms.
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– Toby Bartels
Feb 9 '18 at 21:54
$begingroup$
Indeed Toby and the more standard notation isln
or with an e subscriptlog_e
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– PatrickT
Feb 26 '18 at 6:54
4
$begingroup$
It is standard in analytic number theory to have all logs implicitly natural.
$endgroup$
– Wojowu
Feb 26 '18 at 16:39
add a comment |
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4 Answers
4
active
oldest
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4 Answers
4
active
oldest
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active
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active
oldest
votes
$begingroup$
It is unknown whether
$$
sum_{n=1}^inftyfrac{1}{n^3sin^2n}
$$
converges or not. The difficulty here is that convergence depends on the term $nsin n$ not being too small, which in turn depends on how well $pi$ can be approximated by rational numbers. It is possible that, if $pi$ can be approximated `too well' enough by rationals, then this will diverge. See this MathOverflow question for a discussion of this particular series.
Another even simpler example of a sequence (no summation) for which it is not known whether it converges or not is
$$
x_n=frac{1}{n^2sin n}.
$$
We would expect this to tend to zero, but the proof is beyond what is currently known. Suppose that there were only finitely many rational numbers $p/q$ with $vert p/q-pivertle q^{-3+epsilon}$ (for any $epsilon > 0$), then $x_n$ would tend to zero at rate $O(n^{-epsilon})$. If, on the other hand, there were infinitely many rationals satisfying $vert p/q-pivertle q^{-3-epsilon}$, then infinitely many $x_n$ would be of order at least $n^epsilon$, so it diverges. This can be expressed in terms of the irrationality measure of $pi$. The sequence $x_n$ converges to zero if the irrationality measure of $pi$ is less than 3, and diverges if it is geater than 3. Currently, the best known bound for the irrationality measure is that it is no more than about $7.6063$ (see the link to the mathworld page above). It is expected that the irrationality measure of $pi$ is 2 (it is known that all but a zero-measure set of real numbers have irrationality measure 2). Therefore, it is expected that $x_n$ tends to zero, but there is currently no proof of this.
$endgroup$
$begingroup$
Mentioned at MO mathoverflow.net/questions/100265/…
$endgroup$
– Alexander Chervov
Jun 22 '12 at 13:16
9
$begingroup$
I find it quite incredible that such a simple sequence ($x_n$) has unknown convergence.
$endgroup$
– Anon
Jan 6 '17 at 2:32
1
$begingroup$
Also mentioned on Wikipedia and named “Flint Hills series” both there and in the MO post you referenced.
$endgroup$
– MvG
Jan 6 '17 at 8:47
$begingroup$
I am blown away by this.
$endgroup$
– Randall
Nov 1 '17 at 15:04
add a comment |
$begingroup$
It is unknown whether
$$
sum_{n=1}^inftyfrac{1}{n^3sin^2n}
$$
converges or not. The difficulty here is that convergence depends on the term $nsin n$ not being too small, which in turn depends on how well $pi$ can be approximated by rational numbers. It is possible that, if $pi$ can be approximated `too well' enough by rationals, then this will diverge. See this MathOverflow question for a discussion of this particular series.
Another even simpler example of a sequence (no summation) for which it is not known whether it converges or not is
$$
x_n=frac{1}{n^2sin n}.
$$
We would expect this to tend to zero, but the proof is beyond what is currently known. Suppose that there were only finitely many rational numbers $p/q$ with $vert p/q-pivertle q^{-3+epsilon}$ (for any $epsilon > 0$), then $x_n$ would tend to zero at rate $O(n^{-epsilon})$. If, on the other hand, there were infinitely many rationals satisfying $vert p/q-pivertle q^{-3-epsilon}$, then infinitely many $x_n$ would be of order at least $n^epsilon$, so it diverges. This can be expressed in terms of the irrationality measure of $pi$. The sequence $x_n$ converges to zero if the irrationality measure of $pi$ is less than 3, and diverges if it is geater than 3. Currently, the best known bound for the irrationality measure is that it is no more than about $7.6063$ (see the link to the mathworld page above). It is expected that the irrationality measure of $pi$ is 2 (it is known that all but a zero-measure set of real numbers have irrationality measure 2). Therefore, it is expected that $x_n$ tends to zero, but there is currently no proof of this.
$endgroup$
$begingroup$
Mentioned at MO mathoverflow.net/questions/100265/…
$endgroup$
– Alexander Chervov
Jun 22 '12 at 13:16
9
$begingroup$
I find it quite incredible that such a simple sequence ($x_n$) has unknown convergence.
$endgroup$
– Anon
Jan 6 '17 at 2:32
1
$begingroup$
Also mentioned on Wikipedia and named “Flint Hills series” both there and in the MO post you referenced.
$endgroup$
– MvG
Jan 6 '17 at 8:47
$begingroup$
I am blown away by this.
$endgroup$
– Randall
Nov 1 '17 at 15:04
add a comment |
$begingroup$
It is unknown whether
$$
sum_{n=1}^inftyfrac{1}{n^3sin^2n}
$$
converges or not. The difficulty here is that convergence depends on the term $nsin n$ not being too small, which in turn depends on how well $pi$ can be approximated by rational numbers. It is possible that, if $pi$ can be approximated `too well' enough by rationals, then this will diverge. See this MathOverflow question for a discussion of this particular series.
Another even simpler example of a sequence (no summation) for which it is not known whether it converges or not is
$$
x_n=frac{1}{n^2sin n}.
$$
We would expect this to tend to zero, but the proof is beyond what is currently known. Suppose that there were only finitely many rational numbers $p/q$ with $vert p/q-pivertle q^{-3+epsilon}$ (for any $epsilon > 0$), then $x_n$ would tend to zero at rate $O(n^{-epsilon})$. If, on the other hand, there were infinitely many rationals satisfying $vert p/q-pivertle q^{-3-epsilon}$, then infinitely many $x_n$ would be of order at least $n^epsilon$, so it diverges. This can be expressed in terms of the irrationality measure of $pi$. The sequence $x_n$ converges to zero if the irrationality measure of $pi$ is less than 3, and diverges if it is geater than 3. Currently, the best known bound for the irrationality measure is that it is no more than about $7.6063$ (see the link to the mathworld page above). It is expected that the irrationality measure of $pi$ is 2 (it is known that all but a zero-measure set of real numbers have irrationality measure 2). Therefore, it is expected that $x_n$ tends to zero, but there is currently no proof of this.
$endgroup$
It is unknown whether
$$
sum_{n=1}^inftyfrac{1}{n^3sin^2n}
$$
converges or not. The difficulty here is that convergence depends on the term $nsin n$ not being too small, which in turn depends on how well $pi$ can be approximated by rational numbers. It is possible that, if $pi$ can be approximated `too well' enough by rationals, then this will diverge. See this MathOverflow question for a discussion of this particular series.
Another even simpler example of a sequence (no summation) for which it is not known whether it converges or not is
$$
x_n=frac{1}{n^2sin n}.
$$
We would expect this to tend to zero, but the proof is beyond what is currently known. Suppose that there were only finitely many rational numbers $p/q$ with $vert p/q-pivertle q^{-3+epsilon}$ (for any $epsilon > 0$), then $x_n$ would tend to zero at rate $O(n^{-epsilon})$. If, on the other hand, there were infinitely many rationals satisfying $vert p/q-pivertle q^{-3-epsilon}$, then infinitely many $x_n$ would be of order at least $n^epsilon$, so it diverges. This can be expressed in terms of the irrationality measure of $pi$. The sequence $x_n$ converges to zero if the irrationality measure of $pi$ is less than 3, and diverges if it is geater than 3. Currently, the best known bound for the irrationality measure is that it is no more than about $7.6063$ (see the link to the mathworld page above). It is expected that the irrationality measure of $pi$ is 2 (it is known that all but a zero-measure set of real numbers have irrationality measure 2). Therefore, it is expected that $x_n$ tends to zero, but there is currently no proof of this.
edited Apr 13 '17 at 12:58
Community♦
1
1
answered Feb 5 '11 at 22:51
George LowtherGeorge Lowther
28.9k26094
28.9k26094
$begingroup$
Mentioned at MO mathoverflow.net/questions/100265/…
$endgroup$
– Alexander Chervov
Jun 22 '12 at 13:16
9
$begingroup$
I find it quite incredible that such a simple sequence ($x_n$) has unknown convergence.
$endgroup$
– Anon
Jan 6 '17 at 2:32
1
$begingroup$
Also mentioned on Wikipedia and named “Flint Hills series” both there and in the MO post you referenced.
$endgroup$
– MvG
Jan 6 '17 at 8:47
$begingroup$
I am blown away by this.
$endgroup$
– Randall
Nov 1 '17 at 15:04
add a comment |
$begingroup$
Mentioned at MO mathoverflow.net/questions/100265/…
$endgroup$
– Alexander Chervov
Jun 22 '12 at 13:16
9
$begingroup$
I find it quite incredible that such a simple sequence ($x_n$) has unknown convergence.
$endgroup$
– Anon
Jan 6 '17 at 2:32
1
$begingroup$
Also mentioned on Wikipedia and named “Flint Hills series” both there and in the MO post you referenced.
$endgroup$
– MvG
Jan 6 '17 at 8:47
$begingroup$
I am blown away by this.
$endgroup$
– Randall
Nov 1 '17 at 15:04
$begingroup$
Mentioned at MO mathoverflow.net/questions/100265/…
$endgroup$
– Alexander Chervov
Jun 22 '12 at 13:16
$begingroup$
Mentioned at MO mathoverflow.net/questions/100265/…
$endgroup$
– Alexander Chervov
Jun 22 '12 at 13:16
9
9
$begingroup$
I find it quite incredible that such a simple sequence ($x_n$) has unknown convergence.
$endgroup$
– Anon
Jan 6 '17 at 2:32
$begingroup$
I find it quite incredible that such a simple sequence ($x_n$) has unknown convergence.
$endgroup$
– Anon
Jan 6 '17 at 2:32
1
1
$begingroup$
Also mentioned on Wikipedia and named “Flint Hills series” both there and in the MO post you referenced.
$endgroup$
– MvG
Jan 6 '17 at 8:47
$begingroup$
Also mentioned on Wikipedia and named “Flint Hills series” both there and in the MO post you referenced.
$endgroup$
– MvG
Jan 6 '17 at 8:47
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I am blown away by this.
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– Randall
Nov 1 '17 at 15:04
$begingroup$
I am blown away by this.
$endgroup$
– Randall
Nov 1 '17 at 15:04
add a comment |
$begingroup$
As kind of a joke answer, but technically correct, and motivated by Chandru's deleted reply,
$$sum_{n=0}^infty sin(2pi n!,x)$$
where $x$ is the Euler-Mascheroni constant, or almost any other number whose rationality has not been settled. (If $x$ is rational, the series converges. The implication does not go the other way.)
$endgroup$
1
$begingroup$
Out of curiosity, what makes this a "joke" answer?
$endgroup$
– AlanH
May 17 '13 at 1:58
2
$begingroup$
There is no analysis of decay rate involved.
$endgroup$
– André Nicolas
May 17 '13 at 19:20
$begingroup$
The comma before "or almost any" confused me for a second, better to remove it imho, without it one understands immediately your use of x instead of the usual gamma. (there is an Oxford comma, but not really for "or")
$endgroup$
– PatrickT
Feb 26 '18 at 6:50
add a comment |
$begingroup$
As kind of a joke answer, but technically correct, and motivated by Chandru's deleted reply,
$$sum_{n=0}^infty sin(2pi n!,x)$$
where $x$ is the Euler-Mascheroni constant, or almost any other number whose rationality has not been settled. (If $x$ is rational, the series converges. The implication does not go the other way.)
$endgroup$
1
$begingroup$
Out of curiosity, what makes this a "joke" answer?
$endgroup$
– AlanH
May 17 '13 at 1:58
2
$begingroup$
There is no analysis of decay rate involved.
$endgroup$
– André Nicolas
May 17 '13 at 19:20
$begingroup$
The comma before "or almost any" confused me for a second, better to remove it imho, without it one understands immediately your use of x instead of the usual gamma. (there is an Oxford comma, but not really for "or")
$endgroup$
– PatrickT
Feb 26 '18 at 6:50
add a comment |
$begingroup$
As kind of a joke answer, but technically correct, and motivated by Chandru's deleted reply,
$$sum_{n=0}^infty sin(2pi n!,x)$$
where $x$ is the Euler-Mascheroni constant, or almost any other number whose rationality has not been settled. (If $x$ is rational, the series converges. The implication does not go the other way.)
$endgroup$
As kind of a joke answer, but technically correct, and motivated by Chandru's deleted reply,
$$sum_{n=0}^infty sin(2pi n!,x)$$
where $x$ is the Euler-Mascheroni constant, or almost any other number whose rationality has not been settled. (If $x$ is rational, the series converges. The implication does not go the other way.)
answered Jul 2 '11 at 3:30
André NicolasAndré Nicolas
452k36423808
452k36423808
1
$begingroup$
Out of curiosity, what makes this a "joke" answer?
$endgroup$
– AlanH
May 17 '13 at 1:58
2
$begingroup$
There is no analysis of decay rate involved.
$endgroup$
– André Nicolas
May 17 '13 at 19:20
$begingroup$
The comma before "or almost any" confused me for a second, better to remove it imho, without it one understands immediately your use of x instead of the usual gamma. (there is an Oxford comma, but not really for "or")
$endgroup$
– PatrickT
Feb 26 '18 at 6:50
add a comment |
1
$begingroup$
Out of curiosity, what makes this a "joke" answer?
$endgroup$
– AlanH
May 17 '13 at 1:58
2
$begingroup$
There is no analysis of decay rate involved.
$endgroup$
– André Nicolas
May 17 '13 at 19:20
$begingroup$
The comma before "or almost any" confused me for a second, better to remove it imho, without it one understands immediately your use of x instead of the usual gamma. (there is an Oxford comma, but not really for "or")
$endgroup$
– PatrickT
Feb 26 '18 at 6:50
1
1
$begingroup$
Out of curiosity, what makes this a "joke" answer?
$endgroup$
– AlanH
May 17 '13 at 1:58
$begingroup$
Out of curiosity, what makes this a "joke" answer?
$endgroup$
– AlanH
May 17 '13 at 1:58
2
2
$begingroup$
There is no analysis of decay rate involved.
$endgroup$
– André Nicolas
May 17 '13 at 19:20
$begingroup$
There is no analysis of decay rate involved.
$endgroup$
– André Nicolas
May 17 '13 at 19:20
$begingroup$
The comma before "or almost any" confused me for a second, better to remove it imho, without it one understands immediately your use of x instead of the usual gamma. (there is an Oxford comma, but not really for "or")
$endgroup$
– PatrickT
Feb 26 '18 at 6:50
$begingroup$
The comma before "or almost any" confused me for a second, better to remove it imho, without it one understands immediately your use of x instead of the usual gamma. (there is an Oxford comma, but not really for "or")
$endgroup$
– PatrickT
Feb 26 '18 at 6:50
add a comment |
$begingroup$
It is unknown whether the series:
$$sum_n frac{(-1)^n n}{p_n}$$
converges. Here, $p_n$ is the $n$-th prime number. This problem is posed in Guy's book on unsolved problems in number theory and I am pretty sure that it originated from Erdős.
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3
$begingroup$
Very nice. +1 for Erdős.
$endgroup$
– Bumblebee
Feb 19 '16 at 10:52
$begingroup$
Huh, well, if you accept the prime number theorem, then this converges, else, it is unknowable.
$endgroup$
– Simply Beautiful Art
Mar 9 '17 at 14:48
$begingroup$
@SimplyBeautifulArt the prime number theorem does not imply convergence of this series.
$endgroup$
– Mustafa Said
Mar 13 '17 at 2:38
$begingroup$
Really? By the alternating test, all I need to do is show that $frac n{p_n}to0$.
$endgroup$
– Simply Beautiful Art
Mar 13 '17 at 12:36
11
$begingroup$
@SimplyBeautifulArt in order to use the alternating series test you must show that the terms of the series are monotonically decreasing.
$endgroup$
– Mustafa Said
Mar 17 '17 at 21:30
add a comment |
$begingroup$
It is unknown whether the series:
$$sum_n frac{(-1)^n n}{p_n}$$
converges. Here, $p_n$ is the $n$-th prime number. This problem is posed in Guy's book on unsolved problems in number theory and I am pretty sure that it originated from Erdős.
$endgroup$
3
$begingroup$
Very nice. +1 for Erdős.
$endgroup$
– Bumblebee
Feb 19 '16 at 10:52
$begingroup$
Huh, well, if you accept the prime number theorem, then this converges, else, it is unknowable.
$endgroup$
– Simply Beautiful Art
Mar 9 '17 at 14:48
$begingroup$
@SimplyBeautifulArt the prime number theorem does not imply convergence of this series.
$endgroup$
– Mustafa Said
Mar 13 '17 at 2:38
$begingroup$
Really? By the alternating test, all I need to do is show that $frac n{p_n}to0$.
$endgroup$
– Simply Beautiful Art
Mar 13 '17 at 12:36
11
$begingroup$
@SimplyBeautifulArt in order to use the alternating series test you must show that the terms of the series are monotonically decreasing.
$endgroup$
– Mustafa Said
Mar 17 '17 at 21:30
add a comment |
$begingroup$
It is unknown whether the series:
$$sum_n frac{(-1)^n n}{p_n}$$
converges. Here, $p_n$ is the $n$-th prime number. This problem is posed in Guy's book on unsolved problems in number theory and I am pretty sure that it originated from Erdős.
$endgroup$
It is unknown whether the series:
$$sum_n frac{(-1)^n n}{p_n}$$
converges. Here, $p_n$ is the $n$-th prime number. This problem is posed in Guy's book on unsolved problems in number theory and I am pretty sure that it originated from Erdős.
edited Aug 9 '17 at 10:50
Mutantoe
604513
604513
answered Feb 22 '14 at 9:41
Mustafa SaidMustafa Said
2,9511913
2,9511913
3
$begingroup$
Very nice. +1 for Erdős.
$endgroup$
– Bumblebee
Feb 19 '16 at 10:52
$begingroup$
Huh, well, if you accept the prime number theorem, then this converges, else, it is unknowable.
$endgroup$
– Simply Beautiful Art
Mar 9 '17 at 14:48
$begingroup$
@SimplyBeautifulArt the prime number theorem does not imply convergence of this series.
$endgroup$
– Mustafa Said
Mar 13 '17 at 2:38
$begingroup$
Really? By the alternating test, all I need to do is show that $frac n{p_n}to0$.
$endgroup$
– Simply Beautiful Art
Mar 13 '17 at 12:36
11
$begingroup$
@SimplyBeautifulArt in order to use the alternating series test you must show that the terms of the series are monotonically decreasing.
$endgroup$
– Mustafa Said
Mar 17 '17 at 21:30
add a comment |
3
$begingroup$
Very nice. +1 for Erdős.
$endgroup$
– Bumblebee
Feb 19 '16 at 10:52
$begingroup$
Huh, well, if you accept the prime number theorem, then this converges, else, it is unknowable.
$endgroup$
– Simply Beautiful Art
Mar 9 '17 at 14:48
$begingroup$
@SimplyBeautifulArt the prime number theorem does not imply convergence of this series.
$endgroup$
– Mustafa Said
Mar 13 '17 at 2:38
$begingroup$
Really? By the alternating test, all I need to do is show that $frac n{p_n}to0$.
$endgroup$
– Simply Beautiful Art
Mar 13 '17 at 12:36
11
$begingroup$
@SimplyBeautifulArt in order to use the alternating series test you must show that the terms of the series are monotonically decreasing.
$endgroup$
– Mustafa Said
Mar 17 '17 at 21:30
3
3
$begingroup$
Very nice. +1 for Erdős.
$endgroup$
– Bumblebee
Feb 19 '16 at 10:52
$begingroup$
Very nice. +1 for Erdős.
$endgroup$
– Bumblebee
Feb 19 '16 at 10:52
$begingroup$
Huh, well, if you accept the prime number theorem, then this converges, else, it is unknowable.
$endgroup$
– Simply Beautiful Art
Mar 9 '17 at 14:48
$begingroup$
Huh, well, if you accept the prime number theorem, then this converges, else, it is unknowable.
$endgroup$
– Simply Beautiful Art
Mar 9 '17 at 14:48
$begingroup$
@SimplyBeautifulArt the prime number theorem does not imply convergence of this series.
$endgroup$
– Mustafa Said
Mar 13 '17 at 2:38
$begingroup$
@SimplyBeautifulArt the prime number theorem does not imply convergence of this series.
$endgroup$
– Mustafa Said
Mar 13 '17 at 2:38
$begingroup$
Really? By the alternating test, all I need to do is show that $frac n{p_n}to0$.
$endgroup$
– Simply Beautiful Art
Mar 13 '17 at 12:36
$begingroup$
Really? By the alternating test, all I need to do is show that $frac n{p_n}to0$.
$endgroup$
– Simply Beautiful Art
Mar 13 '17 at 12:36
11
11
$begingroup$
@SimplyBeautifulArt in order to use the alternating series test you must show that the terms of the series are monotonically decreasing.
$endgroup$
– Mustafa Said
Mar 17 '17 at 21:30
$begingroup$
@SimplyBeautifulArt in order to use the alternating series test you must show that the terms of the series are monotonically decreasing.
$endgroup$
– Mustafa Said
Mar 17 '17 at 21:30
add a comment |
$begingroup$
The Riemann hypothesis is that $sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^{3+epsilon} n}$ converges for any $epsilon > 0$ (see here for a discussion of that $epsilon$).
$endgroup$
3
$begingroup$
Here, $Lambda(n)$ is the logarithm of the prime number $p$ such that $n$ is a power of $p$, if any, or $0$ if $n$ is not a prime power. (It says so in the first link, but I thought it helpful to have it on this page.) I also presume that all of the logarithms are natural logarithms.
$endgroup$
– Toby Bartels
Feb 9 '18 at 21:54
$begingroup$
Indeed Toby and the more standard notation isln
or with an e subscriptlog_e
$endgroup$
– PatrickT
Feb 26 '18 at 6:54
4
$begingroup$
It is standard in analytic number theory to have all logs implicitly natural.
$endgroup$
– Wojowu
Feb 26 '18 at 16:39
add a comment |
$begingroup$
The Riemann hypothesis is that $sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^{3+epsilon} n}$ converges for any $epsilon > 0$ (see here for a discussion of that $epsilon$).
$endgroup$
3
$begingroup$
Here, $Lambda(n)$ is the logarithm of the prime number $p$ such that $n$ is a power of $p$, if any, or $0$ if $n$ is not a prime power. (It says so in the first link, but I thought it helpful to have it on this page.) I also presume that all of the logarithms are natural logarithms.
$endgroup$
– Toby Bartels
Feb 9 '18 at 21:54
$begingroup$
Indeed Toby and the more standard notation isln
or with an e subscriptlog_e
$endgroup$
– PatrickT
Feb 26 '18 at 6:54
4
$begingroup$
It is standard in analytic number theory to have all logs implicitly natural.
$endgroup$
– Wojowu
Feb 26 '18 at 16:39
add a comment |
$begingroup$
The Riemann hypothesis is that $sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^{3+epsilon} n}$ converges for any $epsilon > 0$ (see here for a discussion of that $epsilon$).
$endgroup$
The Riemann hypothesis is that $sum_{n=2}^infty frac{Lambda(n)-1}{n^{1/2}log^{3+epsilon} n}$ converges for any $epsilon > 0$ (see here for a discussion of that $epsilon$).
edited Jan 21 at 18:37
answered Aug 9 '17 at 11:23
reunsreuns
19.9k21147
19.9k21147
3
$begingroup$
Here, $Lambda(n)$ is the logarithm of the prime number $p$ such that $n$ is a power of $p$, if any, or $0$ if $n$ is not a prime power. (It says so in the first link, but I thought it helpful to have it on this page.) I also presume that all of the logarithms are natural logarithms.
$endgroup$
– Toby Bartels
Feb 9 '18 at 21:54
$begingroup$
Indeed Toby and the more standard notation isln
or with an e subscriptlog_e
$endgroup$
– PatrickT
Feb 26 '18 at 6:54
4
$begingroup$
It is standard in analytic number theory to have all logs implicitly natural.
$endgroup$
– Wojowu
Feb 26 '18 at 16:39
add a comment |
3
$begingroup$
Here, $Lambda(n)$ is the logarithm of the prime number $p$ such that $n$ is a power of $p$, if any, or $0$ if $n$ is not a prime power. (It says so in the first link, but I thought it helpful to have it on this page.) I also presume that all of the logarithms are natural logarithms.
$endgroup$
– Toby Bartels
Feb 9 '18 at 21:54
$begingroup$
Indeed Toby and the more standard notation isln
or with an e subscriptlog_e
$endgroup$
– PatrickT
Feb 26 '18 at 6:54
4
$begingroup$
It is standard in analytic number theory to have all logs implicitly natural.
$endgroup$
– Wojowu
Feb 26 '18 at 16:39
3
3
$begingroup$
Here, $Lambda(n)$ is the logarithm of the prime number $p$ such that $n$ is a power of $p$, if any, or $0$ if $n$ is not a prime power. (It says so in the first link, but I thought it helpful to have it on this page.) I also presume that all of the logarithms are natural logarithms.
$endgroup$
– Toby Bartels
Feb 9 '18 at 21:54
$begingroup$
Here, $Lambda(n)$ is the logarithm of the prime number $p$ such that $n$ is a power of $p$, if any, or $0$ if $n$ is not a prime power. (It says so in the first link, but I thought it helpful to have it on this page.) I also presume that all of the logarithms are natural logarithms.
$endgroup$
– Toby Bartels
Feb 9 '18 at 21:54
$begingroup$
Indeed Toby and the more standard notation is
ln
or with an e subscript log_e
$endgroup$
– PatrickT
Feb 26 '18 at 6:54
$begingroup$
Indeed Toby and the more standard notation is
ln
or with an e subscript log_e
$endgroup$
– PatrickT
Feb 26 '18 at 6:54
4
4
$begingroup$
It is standard in analytic number theory to have all logs implicitly natural.
$endgroup$
– Wojowu
Feb 26 '18 at 16:39
$begingroup$
It is standard in analytic number theory to have all logs implicitly natural.
$endgroup$
– Wojowu
Feb 26 '18 at 16:39
add a comment |
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35
$begingroup$
Let $a_n = 1$ if $p_n+2$ is a prime where $p_n$ is the $n$th prime. The convergence of $sum_n a_n$ is equivalent to the twin prime conjecture. Of course, this isn't the likely sense in which the OP asked the question but maybe the question needs to be more specific?
$endgroup$
– Dinesh
Feb 5 '11 at 17:36
12
$begingroup$
The problem with this question is that one can encode a great deal of other mathematics as the question of whether some artificial series converges, as Dinesh as shown. One has to somehow restrict to a natural series, and it's far from clear what this even means.
$endgroup$
– Qiaochu Yuan
Feb 5 '11 at 17:44
$begingroup$
@Dinesh: In fact any statement on proving something is countably infinite could be restated using an indicator function like what you have stated. Anyway good one.
$endgroup$
– user17762
Feb 5 '11 at 17:47
2
$begingroup$
Thanks guys, all good points. I clarified things a bit. (My question reminds me of the Ali G interview where he asks "Will a computer ever be able to work out what is 999999999... multiplied by 999999...?".)
$endgroup$
– garageàtrois
Feb 5 '11 at 21:55
1
$begingroup$
The answer to whether a computer can work out... is you are asking for $(10^m-1)*(10^n-1)=10^{(m+n)}-10^m-10^n+1$ and I think Wolfram Alpha can do that already.
$endgroup$
– Ross Millikan
Jul 2 '11 at 3:40