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Concerning this $sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3$ [closed]
$begingroup$
Apparently this sum has this closed form,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248+sqrt{8}left[36G+3pi(3ln{2}-5)-64right]$$
where $G$ is Catalan's constant.
Is this correct?
sequences-and-series binomial-coefficients
edited Jan 21 at 10:07
Tito Piezas III
27.7k367176
asked Jan 21 at 9:43
user583851user583851
508110
$endgroup$
closed as off-topic by mrtaurho, Claude Leibovici, Jose Arnaldo Bebita Dris, choco_addicted, Adrian Keister Jan 21 at 14:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Claude Leibovici, Jose Arnaldo Bebita Dris, choco_addicted, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Apparently this sum has this closed form,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248+sqrt{8}left[36G+3pi(3ln{2}-5)-64right]$$
where $G$ is Catalan's constant.
Is this correct?
sequences-and-series binomial-coefficients
edited Jan 21 at 10:07
Tito Piezas III
27.7k367176
asked Jan 21 at 9:43
user583851user583851
508110
$endgroup$
closed as off-topic by mrtaurho, Claude Leibovici, Jose Arnaldo Bebita Dris, choco_addicted, Adrian Keister Jan 21 at 14:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Claude Leibovici, Jose Arnaldo Bebita Dris, choco_addicted, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Where $G$ is what?
$endgroup$
– José Carlos Santos
Jan 21 at 9:44
$begingroup$
I am guessing it is Catalan's constant.
$endgroup$
– omegadot
Jan 21 at 9:45
6
$begingroup$
Any context concerning how you came across this sum?
$endgroup$
– omegadot
Jan 21 at 9:48
add a comment |
$begingroup$
Apparently this sum has this closed form,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248+sqrt{8}left[36G+3pi(3ln{2}-5)-64right]$$
where $G$ is Catalan's constant.
Is this correct?
sequences-and-series binomial-coefficients
edited Jan 21 at 10:07
Tito Piezas III
27.7k367176
asked Jan 21 at 9:43
user583851user583851
508110
$endgroup$
Apparently this sum has this closed form,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248+sqrt{8}left[36G+3pi(3ln{2}-5)-64right]$$
where $G$ is Catalan's constant.
Is this correct?
sequences-and-series binomial-coefficients
sequences-and-series binomial-coefficients
edited Jan 21 at 10:07
Tito Piezas III
27.7k367176
asked Jan 21 at 9:43
user583851user583851
508110
edited Jan 21 at 10:07
Tito Piezas III
27.7k367176
asked Jan 21 at 9:43
user583851user583851
508110
edited Jan 21 at 10:07
Tito Piezas III
27.7k367176
edited Jan 21 at 10:07
Tito Piezas III
27.7k367176
edited Jan 21 at 10:07
Tito Piezas III
27.7k367176
27.7k367176
asked Jan 21 at 9:43
user583851user583851
508110
asked Jan 21 at 9:43
user583851user583851
508110
asked Jan 21 at 9:43
user583851user583851
508110
508110
closed as off-topic by mrtaurho, Claude Leibovici, Jose Arnaldo Bebita Dris, choco_addicted, Adrian Keister Jan 21 at 14:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Claude Leibovici, Jose Arnaldo Bebita Dris, choco_addicted, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by mrtaurho, Claude Leibovici, Jose Arnaldo Bebita Dris, choco_addicted, Adrian Keister Jan 21 at 14:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Claude Leibovici, Jose Arnaldo Bebita Dris, choco_addicted, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Where $G$ is what?
$endgroup$
– José Carlos Santos
Jan 21 at 9:44
$begingroup$
I am guessing it is Catalan's constant.
$endgroup$
– omegadot
Jan 21 at 9:45
6
$begingroup$
Any context concerning how you came across this sum?
$endgroup$
– omegadot
Jan 21 at 9:48
add a comment |
2
$begingroup$
Where $G$ is what?
$endgroup$
– José Carlos Santos
Jan 21 at 9:44
$begingroup$
I am guessing it is Catalan's constant.
$endgroup$
– omegadot
Jan 21 at 9:45
6
$begingroup$
Any context concerning how you came across this sum?
$endgroup$
– omegadot
Jan 21 at 9:48
2
2
$begingroup$
Where $G$ is what?
$endgroup$
– José Carlos Santos
Jan 21 at 9:44
$begingroup$
Where $G$ is what?
$endgroup$
– José Carlos Santos
Jan 21 at 9:44
$begingroup$
I am guessing it is Catalan's constant.
$endgroup$
– omegadot
Jan 21 at 9:45
$begingroup$
I am guessing it is Catalan's constant.
$endgroup$
– omegadot
Jan 21 at 9:45
6
6
$begingroup$
Any context concerning how you came across this sum?
$endgroup$
– omegadot
Jan 21 at 9:48
$begingroup$
Any context concerning how you came across this sum?
$endgroup$
– omegadot
Jan 21 at 9:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Mathematica can evaluate your series, hence those more experienced than I can probably derive it from basic principles. We have,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248-128sqrt2-48sqrt2pi+216A+4Btag1$$
where $A,B$ are generalized hypergeometric functions given by,
$$A=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;tfrac12right) = frac{4G+piln2}{4sqrt2}$$
$$B=,_3F_2left(tfrac32,tfrac32,tfrac32;tfrac52,tfrac52;tfrac12right) = frac{9(-4G-piln2+2pi)}{2sqrt2}$$
Substituting $A,B$ into $(1)$, we get,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248+2sqrt{2}left[36G+3pi(3ln{2}-5)-64right]tag2$$
which agrees with your post.
answered Jan 21 at 11:49
Tito Piezas IIITito Piezas III
27.7k367176
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Mathematica can evaluate your series, hence those more experienced than I can probably derive it from basic principles. We have,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248-128sqrt2-48sqrt2pi+216A+4Btag1$$
where $A,B$ are generalized hypergeometric functions given by,
$$A=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;tfrac12right) = frac{4G+piln2}{4sqrt2}$$
$$B=,_3F_2left(tfrac32,tfrac32,tfrac32;tfrac52,tfrac52;tfrac12right) = frac{9(-4G-piln2+2pi)}{2sqrt2}$$
Substituting $A,B$ into $(1)$, we get,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248+2sqrt{2}left[36G+3pi(3ln{2}-5)-64right]tag2$$
which agrees with your post.
answered Jan 21 at 11:49
Tito Piezas IIITito Piezas III
27.7k367176
$endgroup$
add a comment |
$begingroup$
Mathematica can evaluate your series, hence those more experienced than I can probably derive it from basic principles. We have,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248-128sqrt2-48sqrt2pi+216A+4Btag1$$
where $A,B$ are generalized hypergeometric functions given by,
$$A=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;tfrac12right) = frac{4G+piln2}{4sqrt2}$$
$$B=,_3F_2left(tfrac32,tfrac32,tfrac32;tfrac52,tfrac52;tfrac12right) = frac{9(-4G-piln2+2pi)}{2sqrt2}$$
Substituting $A,B$ into $(1)$, we get,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248+2sqrt{2}left[36G+3pi(3ln{2}-5)-64right]tag2$$
which agrees with your post.
answered Jan 21 at 11:49
Tito Piezas IIITito Piezas III
27.7k367176
$endgroup$
add a comment |
$begingroup$
Mathematica can evaluate your series, hence those more experienced than I can probably derive it from basic principles. We have,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248-128sqrt2-48sqrt2pi+216A+4Btag1$$
where $A,B$ are generalized hypergeometric functions given by,
$$A=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;tfrac12right) = frac{4G+piln2}{4sqrt2}$$
$$B=,_3F_2left(tfrac32,tfrac32,tfrac32;tfrac52,tfrac52;tfrac12right) = frac{9(-4G-piln2+2pi)}{2sqrt2}$$
Substituting $A,B$ into $(1)$, we get,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248+2sqrt{2}left[36G+3pi(3ln{2}-5)-64right]tag2$$
which agrees with your post.
answered Jan 21 at 11:49
Tito Piezas IIITito Piezas III
27.7k367176
$endgroup$
Mathematica can evaluate your series, hence those more experienced than I can probably derive it from basic principles. We have,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248-128sqrt2-48sqrt2pi+216A+4Btag1$$
where $A,B$ are generalized hypergeometric functions given by,
$$A=,_3F_2left(tfrac12,tfrac12,tfrac12;tfrac32,tfrac32;tfrac12right) = frac{4G+piln2}{4sqrt2}$$
$$B=,_3F_2left(tfrac32,tfrac32,tfrac32;tfrac52,tfrac52;tfrac12right) = frac{9(-4G-piln2+2pi)}{2sqrt2}$$
Substituting $A,B$ into $(1)$, we get,
$$sum_{n=0}^{infty}{2n+3choose n+1} left(frac{1}{2^n}cdotfrac{3}{2n+1}right)^3=248+2sqrt{2}left[36G+3pi(3ln{2}-5)-64right]tag2$$
which agrees with your post.
answered Jan 21 at 11:49
Tito Piezas IIITito Piezas III
27.7k367176
edited Jan 23 at 5:58
answered Jan 21 at 11:49
Tito Piezas IIITito Piezas III
27.7k367176
answered Jan 21 at 11:49
Tito Piezas IIITito Piezas III
27.7k367176
answered Jan 21 at 11:49
Tito Piezas IIITito Piezas III
27.7k367176
27.7k367176
add a comment |
add a comment |
2
$begingroup$
Where $G$ is what?
$endgroup$
– José Carlos Santos
Jan 21 at 9:44
$begingroup$
I am guessing it is Catalan's constant.
$endgroup$
– omegadot
Jan 21 at 9:45
6
$begingroup$
Any context concerning how you came across this sum?
$endgroup$
– omegadot
Jan 21 at 9:48