Does this infinite primes snake-product converge?
$begingroup$
Form an infinite product of prime ratios as follows.
Start with
$$
frac{2}{3}cdotfrac{7}{5}=frac{14}{15} approx 0.93 ;.
$$
Continue alternating a fraction $< 1$ times the next fraction $>1$,
progressively through the primes:
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}
= frac{2926}{3315} approx 0.88 ;,
$$
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}cdotfrac{23}{29}cdotfrac{37}{31}
=frac{2490026}{2980185} approx 0.83 ;.
$$
Continue this process to $infty$. One way to write the product is
$$
xi = prod_{1,5,9,ldots}^infty
frac{p_i}{p_{i+1}}cdotfrac{p_{i+3}}{p_{i+2}}
$$
where $p_i$ is the $i$-th prime.
I call this the primes snake-product:
My questions are:
Q1. Does the product converge?
Q2. If so, to what value $xi$ does it converge?
Up to the $1$-millionth prime ($15485863$),
the product is about $0.9056$:
Update (26Jan2019): @Peter has calculated out to $p_i=10^{10}$ when the product is
$approx 0.9048$.
sequences-and-series number-theory elementary-number-theory prime-numbers
asked Jan 25 at 11:39
Joseph O'RourkeJoseph O'Rourke
18.2k350112
$endgroup$
|
show 15 more comments
$begingroup$
Form an infinite product of prime ratios as follows.
Start with
$$
frac{2}{3}cdotfrac{7}{5}=frac{14}{15} approx 0.93 ;.
$$
Continue alternating a fraction $< 1$ times the next fraction $>1$,
progressively through the primes:
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}
= frac{2926}{3315} approx 0.88 ;,
$$
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}cdotfrac{23}{29}cdotfrac{37}{31}
=frac{2490026}{2980185} approx 0.83 ;.
$$
Continue this process to $infty$. One way to write the product is
$$
xi = prod_{1,5,9,ldots}^infty
frac{p_i}{p_{i+1}}cdotfrac{p_{i+3}}{p_{i+2}}
$$
where $p_i$ is the $i$-th prime.
I call this the primes snake-product:
My questions are:
Q1. Does the product converge?
Q2. If so, to what value $xi$ does it converge?
Up to the $1$-millionth prime ($15485863$),
the product is about $0.9056$:
Update (26Jan2019): @Peter has calculated out to $p_i=10^{10}$ when the product is
$approx 0.9048$.
sequences-and-series number-theory elementary-number-theory prime-numbers
asked Jan 25 at 11:39
Joseph O'RourkeJoseph O'Rourke
18.2k350112
$endgroup$
3
$begingroup$
Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
$endgroup$
– quarague
Jan 25 at 11:55
2
$begingroup$
$10^8$ th prime : $$0.90482881546cdots $$
$endgroup$
– Peter
Jan 25 at 15:19
2
$begingroup$
You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
$endgroup$
– Haran
Jan 25 at 16:58
2
$begingroup$
@Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
$endgroup$
– Joseph O'Rourke
Jan 25 at 21:20
2
$begingroup$
Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
$endgroup$
– Wojowu
Jan 26 at 11:57
|
show 15 more comments
$begingroup$
Form an infinite product of prime ratios as follows.
Start with
$$
frac{2}{3}cdotfrac{7}{5}=frac{14}{15} approx 0.93 ;.
$$
Continue alternating a fraction $< 1$ times the next fraction $>1$,
progressively through the primes:
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}
= frac{2926}{3315} approx 0.88 ;,
$$
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}cdotfrac{23}{29}cdotfrac{37}{31}
=frac{2490026}{2980185} approx 0.83 ;.
$$
Continue this process to $infty$. One way to write the product is
$$
xi = prod_{1,5,9,ldots}^infty
frac{p_i}{p_{i+1}}cdotfrac{p_{i+3}}{p_{i+2}}
$$
where $p_i$ is the $i$-th prime.
I call this the primes snake-product:
My questions are:
Q1. Does the product converge?
Q2. If so, to what value $xi$ does it converge?
Up to the $1$-millionth prime ($15485863$),
the product is about $0.9056$:
Update (26Jan2019): @Peter has calculated out to $p_i=10^{10}$ when the product is
$approx 0.9048$.
sequences-and-series number-theory elementary-number-theory prime-numbers
asked Jan 25 at 11:39
Joseph O'RourkeJoseph O'Rourke
18.2k350112
$endgroup$
Form an infinite product of prime ratios as follows.
Start with
$$
frac{2}{3}cdotfrac{7}{5}=frac{14}{15} approx 0.93 ;.
$$
Continue alternating a fraction $< 1$ times the next fraction $>1$,
progressively through the primes:
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}
= frac{2926}{3315} approx 0.88 ;,
$$
$$
frac{2}{3}cdotfrac{7}{5}cdotfrac{11}{13}cdotfrac{19}{17}cdotfrac{23}{29}cdotfrac{37}{31}
=frac{2490026}{2980185} approx 0.83 ;.
$$
Continue this process to $infty$. One way to write the product is
$$
xi = prod_{1,5,9,ldots}^infty
frac{p_i}{p_{i+1}}cdotfrac{p_{i+3}}{p_{i+2}}
$$
where $p_i$ is the $i$-th prime.
I call this the primes snake-product:
My questions are:
Q1. Does the product converge?
Q2. If so, to what value $xi$ does it converge?
Up to the $1$-millionth prime ($15485863$),
the product is about $0.9056$:
Update (26Jan2019): @Peter has calculated out to $p_i=10^{10}$ when the product is
$approx 0.9048$.
sequences-and-series number-theory elementary-number-theory prime-numbers
sequences-and-series number-theory elementary-number-theory prime-numbers
asked Jan 25 at 11:39
Joseph O'RourkeJoseph O'Rourke
18.2k350112
asked Jan 25 at 11:39
Joseph O'RourkeJoseph O'Rourke
18.2k350112
edited Jan 26 at 13:26
Joseph O'Rourke
asked Jan 25 at 11:39
Joseph O'RourkeJoseph O'Rourke
18.2k350112
asked Jan 25 at 11:39
Joseph O'RourkeJoseph O'Rourke
18.2k350112
asked Jan 25 at 11:39
Joseph O'RourkeJoseph O'Rourke
18.2k350112
18.2k350112
3
$begingroup$
Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
$endgroup$
– quarague
Jan 25 at 11:55
2
$begingroup$
$10^8$ th prime : $$0.90482881546cdots $$
$endgroup$
– Peter
Jan 25 at 15:19
2
$begingroup$
You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
$endgroup$
– Haran
Jan 25 at 16:58
2
$begingroup$
@Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
$endgroup$
– Joseph O'Rourke
Jan 25 at 21:20
2
$begingroup$
Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
$endgroup$
– Wojowu
Jan 26 at 11:57
|
show 15 more comments
3
$begingroup$
Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
$endgroup$
– quarague
Jan 25 at 11:55
2
$begingroup$
$10^8$ th prime : $$0.90482881546cdots $$
$endgroup$
– Peter
Jan 25 at 15:19
2
$begingroup$
You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
$endgroup$
– Haran
Jan 25 at 16:58
2
$begingroup$
@Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
$endgroup$
– Joseph O'Rourke
Jan 25 at 21:20
2
$begingroup$
Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
$endgroup$
– Wojowu
Jan 26 at 11:57
3
3
$begingroup$
Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
$endgroup$
– quarague
Jan 25 at 11:55
$begingroup$
Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
$endgroup$
– quarague
Jan 25 at 11:55
2
2
$begingroup$
$10^8$ th prime : $$0.90482881546cdots $$
$endgroup$
– Peter
Jan 25 at 15:19
$begingroup$
$10^8$ th prime : $$0.90482881546cdots $$
$endgroup$
– Peter
Jan 25 at 15:19
2
2
$begingroup$
You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
$endgroup$
– Haran
Jan 25 at 16:58
$begingroup$
You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
$endgroup$
– Haran
Jan 25 at 16:58
2
2
$begingroup$
@Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
$endgroup$
– Joseph O'Rourke
Jan 25 at 21:20
$begingroup$
@Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
$endgroup$
– Joseph O'Rourke
Jan 25 at 21:20
2
2
$begingroup$
Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
$endgroup$
– Wojowu
Jan 26 at 11:57
$begingroup$
Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
$endgroup$
– Wojowu
Jan 26 at 11:57
|
show 15 more comments
1 Answer
1
active
oldest
votes
$begingroup$
A very extended comment explaining why this problem is probably difficult.
Let $g_n=p_{2n}-p_{2n-1}$. The product we are looking at is then
$$prod_{n=1}^inftyleft(1-frac{g_n}{p_{2n}}right)^{(-1)^n}.$$
Taking logarithms, we are met with a sum of the form
$$sum_{n=1}^inftyleft((-1)^{n-1}frac{g_n}{p_{2n}}+Oleft(left(frac{g_n}{p_{2n}}right)^2right)right).$$
Using results due to Heath-Brown on second moments on prime gaps (see here), namely $sum_{k=1}^ng_k^2=O(x^{7/6+varepsilon})$, by summation by parts we can bound the sum of the error terms by a finite value.
Hence we are left with an alternating sum of $g_n/p_{2n}$. To deal with this, we essentially have to show the sums $sum_{n=1}^N(-1)^{n-1}g_n$ are asymptotically smaller than $p_{2N}$ (this won't guarantee convergence, but we definitely want that to hold). Using the notation of this MO answer (and the paper it cites), this sum is equal to $S(2N;1,4)-S(2N;3,4)$. What we would like to know is that this difference is $o(p_{2N})$. So you see we are quickly lead to investigating asymptotics of $S(N;a,q)$. Conjecturally, we have
$$S(N;a,q)simfrac{p_N}{q}$$
(so that the gaps are in some sense equidistributed), but available bounds are much weaker. In the cases we are interested in, we only get
$$liminffrac{S(N;a,4)}{p_N}geqfrac{1}{256}$$
unconditionally, and even conditionally on prime tuples conjecture we get $geq 1/32$, while what we would like is for the limits to exist and be equal.
Hence, as you can see, the available methods are not capable of showing that the difference $S(2N;1,4)-S(2N;3,4)$ is asymptotically small.
answered Jan 26 at 12:46
WojowuWojowu
18.9k23173
$endgroup$
1
$begingroup$
Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
$endgroup$
– Joseph O'Rourke
Jan 26 at 13:10
1
$begingroup$
@JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
$endgroup$
– Wojowu
Jan 26 at 13:29
3
$begingroup$
One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
$endgroup$
– reuns
Jan 27 at 21:44
add a comment |
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$begingroup$
A very extended comment explaining why this problem is probably difficult.
Let $g_n=p_{2n}-p_{2n-1}$. The product we are looking at is then
$$prod_{n=1}^inftyleft(1-frac{g_n}{p_{2n}}right)^{(-1)^n}.$$
Taking logarithms, we are met with a sum of the form
$$sum_{n=1}^inftyleft((-1)^{n-1}frac{g_n}{p_{2n}}+Oleft(left(frac{g_n}{p_{2n}}right)^2right)right).$$
Using results due to Heath-Brown on second moments on prime gaps (see here), namely $sum_{k=1}^ng_k^2=O(x^{7/6+varepsilon})$, by summation by parts we can bound the sum of the error terms by a finite value.
Hence we are left with an alternating sum of $g_n/p_{2n}$. To deal with this, we essentially have to show the sums $sum_{n=1}^N(-1)^{n-1}g_n$ are asymptotically smaller than $p_{2N}$ (this won't guarantee convergence, but we definitely want that to hold). Using the notation of this MO answer (and the paper it cites), this sum is equal to $S(2N;1,4)-S(2N;3,4)$. What we would like to know is that this difference is $o(p_{2N})$. So you see we are quickly lead to investigating asymptotics of $S(N;a,q)$. Conjecturally, we have
$$S(N;a,q)simfrac{p_N}{q}$$
(so that the gaps are in some sense equidistributed), but available bounds are much weaker. In the cases we are interested in, we only get
$$liminffrac{S(N;a,4)}{p_N}geqfrac{1}{256}$$
unconditionally, and even conditionally on prime tuples conjecture we get $geq 1/32$, while what we would like is for the limits to exist and be equal.
Hence, as you can see, the available methods are not capable of showing that the difference $S(2N;1,4)-S(2N;3,4)$ is asymptotically small.
answered Jan 26 at 12:46
WojowuWojowu
18.9k23173
$endgroup$
1
$begingroup$
Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
$endgroup$
– Joseph O'Rourke
Jan 26 at 13:10
1
$begingroup$
@JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
$endgroup$
– Wojowu
Jan 26 at 13:29
3
$begingroup$
One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
$endgroup$
– reuns
Jan 27 at 21:44
add a comment |
$begingroup$
A very extended comment explaining why this problem is probably difficult.
Let $g_n=p_{2n}-p_{2n-1}$. The product we are looking at is then
$$prod_{n=1}^inftyleft(1-frac{g_n}{p_{2n}}right)^{(-1)^n}.$$
Taking logarithms, we are met with a sum of the form
$$sum_{n=1}^inftyleft((-1)^{n-1}frac{g_n}{p_{2n}}+Oleft(left(frac{g_n}{p_{2n}}right)^2right)right).$$
Using results due to Heath-Brown on second moments on prime gaps (see here), namely $sum_{k=1}^ng_k^2=O(x^{7/6+varepsilon})$, by summation by parts we can bound the sum of the error terms by a finite value.
Hence we are left with an alternating sum of $g_n/p_{2n}$. To deal with this, we essentially have to show the sums $sum_{n=1}^N(-1)^{n-1}g_n$ are asymptotically smaller than $p_{2N}$ (this won't guarantee convergence, but we definitely want that to hold). Using the notation of this MO answer (and the paper it cites), this sum is equal to $S(2N;1,4)-S(2N;3,4)$. What we would like to know is that this difference is $o(p_{2N})$. So you see we are quickly lead to investigating asymptotics of $S(N;a,q)$. Conjecturally, we have
$$S(N;a,q)simfrac{p_N}{q}$$
(so that the gaps are in some sense equidistributed), but available bounds are much weaker. In the cases we are interested in, we only get
$$liminffrac{S(N;a,4)}{p_N}geqfrac{1}{256}$$
unconditionally, and even conditionally on prime tuples conjecture we get $geq 1/32$, while what we would like is for the limits to exist and be equal.
Hence, as you can see, the available methods are not capable of showing that the difference $S(2N;1,4)-S(2N;3,4)$ is asymptotically small.
answered Jan 26 at 12:46
WojowuWojowu
18.9k23173
$endgroup$
1
$begingroup$
Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
$endgroup$
– Joseph O'Rourke
Jan 26 at 13:10
1
$begingroup$
@JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
$endgroup$
– Wojowu
Jan 26 at 13:29
3
$begingroup$
One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
$endgroup$
– reuns
Jan 27 at 21:44
add a comment |
$begingroup$
A very extended comment explaining why this problem is probably difficult.
Let $g_n=p_{2n}-p_{2n-1}$. The product we are looking at is then
$$prod_{n=1}^inftyleft(1-frac{g_n}{p_{2n}}right)^{(-1)^n}.$$
Taking logarithms, we are met with a sum of the form
$$sum_{n=1}^inftyleft((-1)^{n-1}frac{g_n}{p_{2n}}+Oleft(left(frac{g_n}{p_{2n}}right)^2right)right).$$
Using results due to Heath-Brown on second moments on prime gaps (see here), namely $sum_{k=1}^ng_k^2=O(x^{7/6+varepsilon})$, by summation by parts we can bound the sum of the error terms by a finite value.
Hence we are left with an alternating sum of $g_n/p_{2n}$. To deal with this, we essentially have to show the sums $sum_{n=1}^N(-1)^{n-1}g_n$ are asymptotically smaller than $p_{2N}$ (this won't guarantee convergence, but we definitely want that to hold). Using the notation of this MO answer (and the paper it cites), this sum is equal to $S(2N;1,4)-S(2N;3,4)$. What we would like to know is that this difference is $o(p_{2N})$. So you see we are quickly lead to investigating asymptotics of $S(N;a,q)$. Conjecturally, we have
$$S(N;a,q)simfrac{p_N}{q}$$
(so that the gaps are in some sense equidistributed), but available bounds are much weaker. In the cases we are interested in, we only get
$$liminffrac{S(N;a,4)}{p_N}geqfrac{1}{256}$$
unconditionally, and even conditionally on prime tuples conjecture we get $geq 1/32$, while what we would like is for the limits to exist and be equal.
Hence, as you can see, the available methods are not capable of showing that the difference $S(2N;1,4)-S(2N;3,4)$ is asymptotically small.
answered Jan 26 at 12:46
WojowuWojowu
18.9k23173
$endgroup$
A very extended comment explaining why this problem is probably difficult.
Let $g_n=p_{2n}-p_{2n-1}$. The product we are looking at is then
$$prod_{n=1}^inftyleft(1-frac{g_n}{p_{2n}}right)^{(-1)^n}.$$
Taking logarithms, we are met with a sum of the form
$$sum_{n=1}^inftyleft((-1)^{n-1}frac{g_n}{p_{2n}}+Oleft(left(frac{g_n}{p_{2n}}right)^2right)right).$$
Using results due to Heath-Brown on second moments on prime gaps (see here), namely $sum_{k=1}^ng_k^2=O(x^{7/6+varepsilon})$, by summation by parts we can bound the sum of the error terms by a finite value.
Hence we are left with an alternating sum of $g_n/p_{2n}$. To deal with this, we essentially have to show the sums $sum_{n=1}^N(-1)^{n-1}g_n$ are asymptotically smaller than $p_{2N}$ (this won't guarantee convergence, but we definitely want that to hold). Using the notation of this MO answer (and the paper it cites), this sum is equal to $S(2N;1,4)-S(2N;3,4)$. What we would like to know is that this difference is $o(p_{2N})$. So you see we are quickly lead to investigating asymptotics of $S(N;a,q)$. Conjecturally, we have
$$S(N;a,q)simfrac{p_N}{q}$$
(so that the gaps are in some sense equidistributed), but available bounds are much weaker. In the cases we are interested in, we only get
$$liminffrac{S(N;a,4)}{p_N}geqfrac{1}{256}$$
unconditionally, and even conditionally on prime tuples conjecture we get $geq 1/32$, while what we would like is for the limits to exist and be equal.
Hence, as you can see, the available methods are not capable of showing that the difference $S(2N;1,4)-S(2N;3,4)$ is asymptotically small.
answered Jan 26 at 12:46
WojowuWojowu
18.9k23173
edited Jan 26 at 16:15
answered Jan 26 at 12:46
WojowuWojowu
18.9k23173
answered Jan 26 at 12:46
WojowuWojowu
18.9k23173
answered Jan 26 at 12:46
WojowuWojowu
18.9k23173
18.9k23173
1
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Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
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– Joseph O'Rourke
Jan 26 at 13:10
1
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@JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
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– Wojowu
Jan 26 at 13:29
3
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One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
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– reuns
Jan 27 at 21:44
add a comment |
1
$begingroup$
Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
$endgroup$
– Joseph O'Rourke
Jan 26 at 13:10
1
$begingroup$
@JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
$endgroup$
– Wojowu
Jan 26 at 13:29
3
$begingroup$
One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
$endgroup$
– reuns
Jan 27 at 21:44
1
1
$begingroup$
Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
$endgroup$
– Joseph O'Rourke
Jan 26 at 13:10
$begingroup$
Thank you for your knowledgeable remarks. May I ask: Do you know if there a well-known conjecture that would imply the snake-product converges?
$endgroup$
– Joseph O'Rourke
Jan 26 at 13:10
1
1
$begingroup$
@JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
$endgroup$
– Wojowu
Jan 26 at 13:29
$begingroup$
@JosephO'Rourke Not really, no. It should follow from suitable error term estimates for $S(N;a,q)$, but I'm not aware of such conjectures. I haven't exactly checked the references in the linked answer, you might want to look there.
$endgroup$
– Wojowu
Jan 26 at 13:29
3
3
$begingroup$
One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
$endgroup$
– reuns
Jan 27 at 21:44
$begingroup$
One of the strongest conjecture for the prime gap is that $g(n),n in [N,2N]$ looks like a random variable $G_N$ with $P(G_N > alog N) = e^{-a} log N$ and you are essentially asking $g(n)$ to be at least slightly independent in $n$ @JosephO'Rourke
$endgroup$
– reuns
Jan 27 at 21:44
add a comment |
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$begingroup$
Using the prime number theorem as an approximation, if you set $p_n=nln(n)$, does your product converge?
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– quarague
Jan 25 at 11:55
2
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$10^8$ th prime : $$0.90482881546cdots $$
$endgroup$
– Peter
Jan 25 at 15:19
2
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You could prove that the value of $prod_{i=1}^{infty} frac{p_{2n-1}}{p_{2n}} = P$ converges. Then, you can easily see that $P < xi < frac{1}{P}$, which would show that your product converges.
$endgroup$
– Haran
Jan 25 at 16:58
2
$begingroup$
@Haran: But perhaps $P=0$, and I'd be left with $0 < xi < infty$.
$endgroup$
– Joseph O'Rourke
Jan 25 at 21:20
2
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Since (as I note above) the products of $frac{p_i}{p_{i+1}}$ and of $frac{p_{i+3}}{p_{i+2}}$ do not converge absolutely, we are lead to considering cancelations between the terms. This is closely related to asymptotics of alternating sums of primes, of which we have only a limited understanding. See this MO question. I will elaborate in an answer.
$endgroup$
– Wojowu
Jan 26 at 11:57