Mixing
About this sum $sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}f(n)$ [closed]
$begingroup$
Observing this sum
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}f(n)$$ we got these two results:
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}cdot frac{n}{(2n-1)^2}=frac{8pi}{Gamma^4left(frac{1}{4}right)}tag1$$
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}cdot frac{1}{(2n-1)^3}=-frac{48pi}{Gamma^4left(frac{1}{4}right)}tag2$$
How can we prove them?
sequences-and-series
asked Jan 21 at 20:22
user583851user583851
508110
$endgroup$
closed as off-topic by mrtaurho, max_zorn, José Carlos Santos, rtybase, Did Jan 27 at 12:12
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, max_zorn, José Carlos Santos, rtybase, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Observing this sum
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}f(n)$$ we got these two results:
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}cdot frac{n}{(2n-1)^2}=frac{8pi}{Gamma^4left(frac{1}{4}right)}tag1$$
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}cdot frac{1}{(2n-1)^3}=-frac{48pi}{Gamma^4left(frac{1}{4}right)}tag2$$
How can we prove them?
sequences-and-series
asked Jan 21 at 20:22
user583851user583851
508110
$endgroup$
closed as off-topic by mrtaurho, max_zorn, José Carlos Santos, rtybase, Did Jan 27 at 12:12
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, max_zorn, José Carlos Santos, rtybase, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Clausen's formula for the square of a 2F1, or FL expansions.
$endgroup$
– Jack D'Aurizio
Jan 21 at 20:44
add a comment |
$begingroup$
Observing this sum
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}f(n)$$ we got these two results:
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}cdot frac{n}{(2n-1)^2}=frac{8pi}{Gamma^4left(frac{1}{4}right)}tag1$$
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}cdot frac{1}{(2n-1)^3}=-frac{48pi}{Gamma^4left(frac{1}{4}right)}tag2$$
How can we prove them?
sequences-and-series
asked Jan 21 at 20:22
user583851user583851
508110
$endgroup$
Observing this sum
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}f(n)$$ we got these two results:
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}cdot frac{n}{(2n-1)^2}=frac{8pi}{Gamma^4left(frac{1}{4}right)}tag1$$
$$sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}cdot frac{1}{(2n-1)^3}=-frac{48pi}{Gamma^4left(frac{1}{4}right)}tag2$$
How can we prove them?
sequences-and-series
sequences-and-series
asked Jan 21 at 20:22
user583851user583851
508110
asked Jan 21 at 20:22
user583851user583851
508110
asked Jan 21 at 20:22
user583851user583851
508110
asked Jan 21 at 20:22
user583851user583851
508110
asked Jan 21 at 20:22
user583851user583851
508110
508110
closed as off-topic by mrtaurho, max_zorn, José Carlos Santos, rtybase, Did Jan 27 at 12:12
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, max_zorn, José Carlos Santos, rtybase, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by mrtaurho, max_zorn, José Carlos Santos, rtybase, Did Jan 27 at 12:12
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, max_zorn, José Carlos Santos, rtybase, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Clausen's formula for the square of a 2F1, or FL expansions.
$endgroup$
– Jack D'Aurizio
Jan 21 at 20:44
add a comment |
1
$begingroup$
Clausen's formula for the square of a 2F1, or FL expansions.
$endgroup$
– Jack D'Aurizio
Jan 21 at 20:44
1
1
$begingroup$
Clausen's formula for the square of a 2F1, or FL expansions.
$endgroup$
– Jack D'Aurizio
Jan 21 at 20:44
$begingroup$
Clausen's formula for the square of a 2F1, or FL expansions.
$endgroup$
– Jack D'Aurizio
Jan 21 at 20:44
add a comment |
0
active
oldest
votes
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Clausen's formula for the square of a 2F1, or FL expansions.
$endgroup$
– Jack D'Aurizio
Jan 21 at 20:44