About this sum $sum_{n=0}^{infty}frac{{2n choose n}^3}{64^n}f(n)$ [closed]












3












$begingroup$


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







share|cite|improve this question









$endgroup$



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


















3












$begingroup$


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







share|cite|improve this question









$endgroup$



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
















3












3








3





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
















  • 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












0






active

oldest

votes

















0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes

Popular posts from this blog

Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

Image
28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
Read more

Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

Image
up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
Read more

A Topological Invariant for $pi_3(U(n))$

Image
3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
Read more