Value of $prod_{n=2}^{infty} frac{n^2+1}{n^2-1}$
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Consider the following product:
$$prod_{n=2}^{infty} frac{n^2+1}{n^2-1} = prod_{n=2}^{infty} frac{1+frac{1}{n}}{1-frac{1}{n}}approx 3.67608...$$
It seems to be close to OEIS A156648, i.e. $prod_{n=1}^{infty} 1+frac{1}{n^2}$, which is also expressible as $frac{sinh(pi)}{pi}$.
Is the first product expressible in similar closed form?
products
asked Nov 19 at 20:11
g3nuine
1188
add a comment |
up vote
0
down vote
favorite
Consider the following product:
$$prod_{n=2}^{infty} frac{n^2+1}{n^2-1} = prod_{n=2}^{infty} frac{1+frac{1}{n}}{1-frac{1}{n}}approx 3.67608...$$
It seems to be close to OEIS A156648, i.e. $prod_{n=1}^{infty} 1+frac{1}{n^2}$, which is also expressible as $frac{sinh(pi)}{pi}$.
Is the first product expressible in similar closed form?
products
asked Nov 19 at 20:11
g3nuine
1188
That's only the partial product form (just checked myself). OP presumably wants an exact value for the product itself - hypothetically you could take $m rightarrow infty$, but that sems messy with four different gamma functions
– Eevee Trainer
Nov 19 at 20:16
1
@g3nuine Note that we indeed have $prod_{n=1}^{infty} left(1+frac{1}{n^2}right)=frac{sinh(pi)}{pi}$, as given in your OEIS link, but $$prod_{n=2}^{infty} left(1+frac{1}{n^2}right)=frac{1}{1+frac{1}{1^2}}cdot prod_{n=1}^{infty}left(1+frac{1}{n^2}right)=frac{sinh(pi)}{2pi}$$
– projectilemotion
Nov 19 at 20:26
@projectilemotion You're right my bad, copy paste error. I've edited it
– g3nuine
Nov 19 at 20:30
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the following product:
$$prod_{n=2}^{infty} frac{n^2+1}{n^2-1} = prod_{n=2}^{infty} frac{1+frac{1}{n}}{1-frac{1}{n}}approx 3.67608...$$
It seems to be close to OEIS A156648, i.e. $prod_{n=1}^{infty} 1+frac{1}{n^2}$, which is also expressible as $frac{sinh(pi)}{pi}$.
Is the first product expressible in similar closed form?
products
asked Nov 19 at 20:11
g3nuine
1188
Consider the following product:
$$prod_{n=2}^{infty} frac{n^2+1}{n^2-1} = prod_{n=2}^{infty} frac{1+frac{1}{n}}{1-frac{1}{n}}approx 3.67608...$$
It seems to be close to OEIS A156648, i.e. $prod_{n=1}^{infty} 1+frac{1}{n^2}$, which is also expressible as $frac{sinh(pi)}{pi}$.
Is the first product expressible in similar closed form?
products
products
asked Nov 19 at 20:11
g3nuine
1188
asked Nov 19 at 20:11
g3nuine
1188
edited Nov 19 at 20:29
asked Nov 19 at 20:11
g3nuine
1188
asked Nov 19 at 20:11
g3nuine
1188
asked Nov 19 at 20:11
g3nuine
1188
1188
That's only the partial product form (just checked myself). OP presumably wants an exact value for the product itself - hypothetically you could take $m rightarrow infty$, but that sems messy with four different gamma functions
– Eevee Trainer
Nov 19 at 20:16
1
@g3nuine Note that we indeed have $prod_{n=1}^{infty} left(1+frac{1}{n^2}right)=frac{sinh(pi)}{pi}$, as given in your OEIS link, but $$prod_{n=2}^{infty} left(1+frac{1}{n^2}right)=frac{1}{1+frac{1}{1^2}}cdot prod_{n=1}^{infty}left(1+frac{1}{n^2}right)=frac{sinh(pi)}{2pi}$$
– projectilemotion
Nov 19 at 20:26
@projectilemotion You're right my bad, copy paste error. I've edited it
– g3nuine
Nov 19 at 20:30
add a comment |
That's only the partial product form (just checked myself). OP presumably wants an exact value for the product itself - hypothetically you could take $m rightarrow infty$, but that sems messy with four different gamma functions
– Eevee Trainer
Nov 19 at 20:16
1
@g3nuine Note that we indeed have $prod_{n=1}^{infty} left(1+frac{1}{n^2}right)=frac{sinh(pi)}{pi}$, as given in your OEIS link, but $$prod_{n=2}^{infty} left(1+frac{1}{n^2}right)=frac{1}{1+frac{1}{1^2}}cdot prod_{n=1}^{infty}left(1+frac{1}{n^2}right)=frac{sinh(pi)}{2pi}$$
– projectilemotion
Nov 19 at 20:26
@projectilemotion You're right my bad, copy paste error. I've edited it
– g3nuine
Nov 19 at 20:30
That's only the partial product form (just checked myself). OP presumably wants an exact value for the product itself - hypothetically you could take $m rightarrow infty$, but that sems messy with four different gamma functions
– Eevee Trainer
Nov 19 at 20:16
That's only the partial product form (just checked myself). OP presumably wants an exact value for the product itself - hypothetically you could take $m rightarrow infty$, but that sems messy with four different gamma functions
– Eevee Trainer
Nov 19 at 20:16
1
1
@g3nuine Note that we indeed have $prod_{n=1}^{infty} left(1+frac{1}{n^2}right)=frac{sinh(pi)}{pi}$, as given in your OEIS link, but $$prod_{n=2}^{infty} left(1+frac{1}{n^2}right)=frac{1}{1+frac{1}{1^2}}cdot prod_{n=1}^{infty}left(1+frac{1}{n^2}right)=frac{sinh(pi)}{2pi}$$
– projectilemotion
Nov 19 at 20:26
@g3nuine Note that we indeed have $prod_{n=1}^{infty} left(1+frac{1}{n^2}right)=frac{sinh(pi)}{pi}$, as given in your OEIS link, but $$prod_{n=2}^{infty} left(1+frac{1}{n^2}right)=frac{1}{1+frac{1}{1^2}}cdot prod_{n=1}^{infty}left(1+frac{1}{n^2}right)=frac{sinh(pi)}{2pi}$$
– projectilemotion
Nov 19 at 20:26
@projectilemotion You're right my bad, copy paste error. I've edited it
– g3nuine
Nov 19 at 20:30
@projectilemotion You're right my bad, copy paste error. I've edited it
– g3nuine
Nov 19 at 20:30
add a comment |
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
Write your product as
$$prod_{n=2}^infty frac{1+frac{1}{n^2}}{1-frac{1}{n^2}}$$
and then notice that the product of the denominators telescopes:
$$prod_{n=2}^infty bigg(1-frac{1}{n^2}bigg)=prod_{n=2}^infty frac{(n+1)(n-1)}{n^2}=frac{1}{2}$$
making your product equal to
$$2prod_{n=2}^infty bigg(1+frac{1}{n^2}bigg)=frac{sinh(pi)}{pi}$$
answered Nov 19 at 20:20
Frpzzd
20.2k638104
Hi, I believe your answer is currently incorrect. $2sinh(pi)/pi approx 7.35$; however, Wolfram Alpha says that the product equals $3.676ldots$ Seems like you're off by a factor of $2$.
– Ekesh
Nov 19 at 20:21
1
@Ekesh Ah, this is because the OP's closed form should be $sinh(pi)/2pi$. I will fix it.
– Frpzzd
Nov 19 at 20:25
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Write your product as
$$prod_{n=2}^infty frac{1+frac{1}{n^2}}{1-frac{1}{n^2}}$$
and then notice that the product of the denominators telescopes:
$$prod_{n=2}^infty bigg(1-frac{1}{n^2}bigg)=prod_{n=2}^infty frac{(n+1)(n-1)}{n^2}=frac{1}{2}$$
making your product equal to
$$2prod_{n=2}^infty bigg(1+frac{1}{n^2}bigg)=frac{sinh(pi)}{pi}$$
answered Nov 19 at 20:20
Frpzzd
20.2k638104
Hi, I believe your answer is currently incorrect. $2sinh(pi)/pi approx 7.35$; however, Wolfram Alpha says that the product equals $3.676ldots$ Seems like you're off by a factor of $2$.
– Ekesh
Nov 19 at 20:21
1
@Ekesh Ah, this is because the OP's closed form should be $sinh(pi)/2pi$. I will fix it.
– Frpzzd
Nov 19 at 20:25
add a comment |
up vote
5
down vote
accepted
Write your product as
$$prod_{n=2}^infty frac{1+frac{1}{n^2}}{1-frac{1}{n^2}}$$
and then notice that the product of the denominators telescopes:
$$prod_{n=2}^infty bigg(1-frac{1}{n^2}bigg)=prod_{n=2}^infty frac{(n+1)(n-1)}{n^2}=frac{1}{2}$$
making your product equal to
$$2prod_{n=2}^infty bigg(1+frac{1}{n^2}bigg)=frac{sinh(pi)}{pi}$$
answered Nov 19 at 20:20
Frpzzd
20.2k638104
Hi, I believe your answer is currently incorrect. $2sinh(pi)/pi approx 7.35$; however, Wolfram Alpha says that the product equals $3.676ldots$ Seems like you're off by a factor of $2$.
– Ekesh
Nov 19 at 20:21
1
@Ekesh Ah, this is because the OP's closed form should be $sinh(pi)/2pi$. I will fix it.
– Frpzzd
Nov 19 at 20:25
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Write your product as
$$prod_{n=2}^infty frac{1+frac{1}{n^2}}{1-frac{1}{n^2}}$$
and then notice that the product of the denominators telescopes:
$$prod_{n=2}^infty bigg(1-frac{1}{n^2}bigg)=prod_{n=2}^infty frac{(n+1)(n-1)}{n^2}=frac{1}{2}$$
making your product equal to
$$2prod_{n=2}^infty bigg(1+frac{1}{n^2}bigg)=frac{sinh(pi)}{pi}$$
answered Nov 19 at 20:20
Frpzzd
20.2k638104
Write your product as
$$prod_{n=2}^infty frac{1+frac{1}{n^2}}{1-frac{1}{n^2}}$$
and then notice that the product of the denominators telescopes:
$$prod_{n=2}^infty bigg(1-frac{1}{n^2}bigg)=prod_{n=2}^infty frac{(n+1)(n-1)}{n^2}=frac{1}{2}$$
making your product equal to
$$2prod_{n=2}^infty bigg(1+frac{1}{n^2}bigg)=frac{sinh(pi)}{pi}$$
answered Nov 19 at 20:20
Frpzzd
20.2k638104
edited Nov 19 at 20:25
answered Nov 19 at 20:20
Frpzzd
20.2k638104
answered Nov 19 at 20:20
Frpzzd
20.2k638104
answered Nov 19 at 20:20
Frpzzd
20.2k638104
20.2k638104
Hi, I believe your answer is currently incorrect. $2sinh(pi)/pi approx 7.35$; however, Wolfram Alpha says that the product equals $3.676ldots$ Seems like you're off by a factor of $2$.
– Ekesh
Nov 19 at 20:21
1
@Ekesh Ah, this is because the OP's closed form should be $sinh(pi)/2pi$. I will fix it.
– Frpzzd
Nov 19 at 20:25
add a comment |
Hi, I believe your answer is currently incorrect. $2sinh(pi)/pi approx 7.35$; however, Wolfram Alpha says that the product equals $3.676ldots$ Seems like you're off by a factor of $2$.
– Ekesh
Nov 19 at 20:21
1
@Ekesh Ah, this is because the OP's closed form should be $sinh(pi)/2pi$. I will fix it.
– Frpzzd
Nov 19 at 20:25
Hi, I believe your answer is currently incorrect. $2sinh(pi)/pi approx 7.35$; however, Wolfram Alpha says that the product equals $3.676ldots$ Seems like you're off by a factor of $2$.
– Ekesh
Nov 19 at 20:21
Hi, I believe your answer is currently incorrect. $2sinh(pi)/pi approx 7.35$; however, Wolfram Alpha says that the product equals $3.676ldots$ Seems like you're off by a factor of $2$.
– Ekesh
Nov 19 at 20:21
1
1
@Ekesh Ah, this is because the OP's closed form should be $sinh(pi)/2pi$. I will fix it.
– Frpzzd
Nov 19 at 20:25
@Ekesh Ah, this is because the OP's closed form should be $sinh(pi)/2pi$. I will fix it.
– Frpzzd
Nov 19 at 20:25
add a comment |
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That's only the partial product form (just checked myself). OP presumably wants an exact value for the product itself - hypothetically you could take $m rightarrow infty$, but that sems messy with four different gamma functions
– Eevee Trainer
Nov 19 at 20:16
1
@g3nuine Note that we indeed have $prod_{n=1}^{infty} left(1+frac{1}{n^2}right)=frac{sinh(pi)}{pi}$, as given in your OEIS link, but $$prod_{n=2}^{infty} left(1+frac{1}{n^2}right)=frac{1}{1+frac{1}{1^2}}cdot prod_{n=1}^{infty}left(1+frac{1}{n^2}right)=frac{sinh(pi)}{2pi}$$
– projectilemotion
Nov 19 at 20:26
@projectilemotion You're right my bad, copy paste error. I've edited it
– g3nuine
Nov 19 at 20:30