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EDIT: Context for this investigation can be found in one of my other MO posts, "Pythagorean Theorem for Right-Corner Hyperbolic Simplices?"

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I'm investigating a function that has led me to this series:

$$T(r) = \sum_{k=0}^{\infty}{2k \choose k}\frac{1}{(k+r)^2 16^k}={}_3F_2(\;\;\frac{1}{2},\;r,\;r\;;\;1+r,\;1+r\;;\;\frac{1}{4}\;\;)\frac{1}{r^2}$$

This paper ("A certain series associated with Catalan's constant") by Victor Adamchik spotlights a tantalizingly similar series ...

$$S(r)=\sum_{k=0}^{\infty}{2k\choose k}^2 \frac{1}{(k+r) \;\; 16^k}={}_3F_2(\;\;\frac{1}{2},\;\frac{1}{2},\;r\;;\;1,\;1+r\;;\;1\;\;)\frac{1}{r}$$

... where the square is on the "wrong" factor.

I'm not very familiar with hypergeometric series (yet). Is there an identity that relates $S$ and $T$?

Edit: I've added the "${}_3F_2$" representations of $S$ and $T$.

Edit2: I have a particular interest in the case $r=1/2$:

$$T(1/2) = 4 \cdot {}_3F_2(\;\;\frac{1}{2},\;\frac{1}{2},\;\frac{1}{2}\;;\;\frac{3}{2},\;\frac{3}{2}\;;\;\frac{1}{4}\;\;) = 4 \cdot \Im( Li_2( \exp( \frac{i \pi}{3} ))) = 4 \;\; \sum_{k=1}^{\infty}\frac{\sin(\frac{\pi}{3}k)}{k^2}$$

$$S(1/2)= 2\cdot {}_3F_2(\;\;\frac{1}{2},\;\frac{1}{2},\;\frac{1}{2}\;;\;1,\;\frac{3}{2}\;;\;1\;\;) = \frac{8G}{\pi}$$

where $G$ is the Catalan constant

$$G = \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2} = {}_3 F_2(\;\; \frac{1}{2},\;\frac{1}{2},\;1 \;;\;\frac{3}{2},\;\frac{3}{2}\;;\;-1\;\;)$$

and $Li_2$ is the dilogarithm.

$G$ arises in my analysis of $T$, but the connection isn't clear (to me).

Edit3: Thanks to guidance from the comments, I've expressed $T(1/2)$ in terms of the dilogarithm function. I've also given the "${}_3F_2$" representation of $G$. The connection between $S(1/2)$ (or $G$) and $T(1/2)$ still eludes me, but I've only begun searching through the various caches of identities that have been recommended.

Blue
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    Mathematica says they're both ${}_3 F_2$ functions. I'll take a look, but in the meantime, have you checked http://dlmf.nist.gov/16 and http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/ ? – J. M. isn't a mathematician Nov 09 '10 at 09:38
  • @J. M.: Checking them now ... :) – Blue Nov 09 '10 at 10:02
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    $T(1/2)$ is expressible in terms of dilogarithms: http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/08/05/01/01/07/0001/ – J. M. isn't a mathematician Nov 09 '10 at 15:13
  • From http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/07/01/01/0001/ , we get an integral representation for $T(1/2)$: $\int_0^1 \frac1{u}\arcsin\left(\frac{\sqrt{u}}{2}\right)\mathrm{d}u$. – J. M. isn't a mathematician Nov 09 '10 at 15:20
  • Another integral representation: $2\int_0^{\pi/6}u\cot;u\mathrm{d}u$ – J. M. isn't a mathematician Nov 09 '10 at 15:49
  • does the standard trick of writing $1/(2k+1) = \int_0^1 x^{2k}dx$ help? – Suvrit Nov 09 '10 at 22:27
  • @Suvrit: Interestingly, Adamchik uses that trick in Entry 30 here http://www.cs.cmu.edu/~adamchik/articles/catalan/catalan.htm to get a series for Catalan's constant that's almost like the series for $T(1/2)$, except my $16^k$ is his $8^k$. (So close!) Applying the trick to the $T(1/2)$ series gives an integral that Mathematica immediately evaluates to what is effectively the dilogarithm formula. – Blue Nov 10 '10 at 20:43
  • Yes, I also played around; first you have $16^k$; if you had $(-16)^k$ then things would simplify dramatically I think. – Suvrit Nov 12 '10 at 20:14
  • After days of playing around, I have to agree that it's very tantalizing; I suspect some weird identity for the Lerch transcendent may be the key (since Dirichlet series and the polylogarithms are special cases of the Lerch transcendent). – J. M. isn't a mathematician Nov 14 '10 at 12:10

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