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I. Intro

While trying to solve this post about the function,

$$F(k)=\sum_{n=0}^{\infty}{2n+3\choose n+1} \left(\frac{1}{2^n}\cdot\frac{3}{2n+1}\right)^k$$

for $k=3$, I found out Mathematica can evaluate this in closed-form for general $k$ using the generalized hypergeometric function. Specifically, for $k=3,4$, it involves Catalan's constant $G$ and the "similar" Gieseking's constant $H$,

$$G = \text{Cl}_2\big(\tfrac{\pi}2\big)=\sum_{n=0}^\infty\left(\frac1{(4n+1)^2}-\frac1{(4n+3)^2}\right)=0.91596\dots$$

$$H = \text{Cl}_2\big(\tfrac{\pi}3\big)=\frac{3\sqrt3}4\sum_{n=0}^\infty\left(\frac1{(3n+1)^2}-\frac1{(3n+2)^2}\right)=1.01494\dots$$

with Clausen's integral $\text{Cl}_2(\theta)$.

II. Examples

$$\sum_{n=0}^{\infty}{2n+3\choose n+1} \left(\frac{1}{2^n}\cdot\frac{3}{2n+1}\right)^3=248-128\sqrt2-30\sqrt2\pi+144\alpha$$ $$\sum_{n=0}^{\infty}{2n+3\choose n+1} \left(\frac{1}{2^n}\cdot\frac{3}{2n+1}\right)^4=-1424+720\sqrt3+112\pi+14\pi^3-360H$$

where,

$$\alpha=\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;\color{red}{\tfrac12}\right) = \frac{4G+\pi\ln2}{4\sqrt2}$$

$$H=\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;\color{red}{\tfrac14}\right) =\text{Cl}_2\big(\tfrac{\pi}3\big)$$

III. Notes

  1. The constants $G$ and $H$ seem to share certain features. As $\delta=\exp\big(\frac{2G}\pi\big)$ and $\beta=\exp\big(\frac{H}\pi\big)$ they are the Kneser-Mahler polynomial constants. (See p. 231 of "Mathematical Constants" by S. Finch.)
  2. $G$ is a rational multiple of the volume of an ideal hyperbolic octahedron, while $H$ is the volume of the hyperbolic Gieseking manifold.
  3. And so on.

IV. Question

Q: Are there other examples of a series or function $P(k)$ such that it has a closed-form in terms of Catalan's constant $G$ or Gieseking's constant $H$ depending on $k$?

Tito Piezas III
  • 54,565
  • Note: $\enspace$ I think it’s better to discuss the integral of the last part of the equation chain $\enspace$$\displaystyle {}{m+2}F{m+1}\left(\frac{1}{2}…\frac{1}{2};\frac{3}{2}…\frac{3}{2};\frac{4}{e^{2x}}\right) = \sum\limits_{n=0}^\infty\frac{1}{e^{2xn}(2n+1)^{m+1}},$$\displaystyle = \frac{e^x}{m!} \sum\limits_{v=0}^m {\binom m v }(-x)^{m-v} \int\limits_x^\infty\frac{t^v~dt}{\sqrt{e^{2t}-4}},$ . Perhaps exists literature about $\displaystyle \int\limits_z^\infty t^a\sqrt{\frac{t}{e^t-1}}dt,$ which could help. – user90369 Jan 28 '19 at 12:58
  • @user90369: In general, $\displaystyle\int_0^\infty\frac{x^{n-1}}{e^x-a}~dx ~=~ \frac{\Gamma(n)\cdot\text{Li}_a(n)}a,~$ so evaluating the integral you suggest might involve fractional calculus, incomplete $\Gamma$ functions, and polylogarithms. – Lucian May 02 '19 at 20:03
  • In my note above is a mistake, the middle series must be $\sum\limits_{n=0}^\infty {\binom{2n} n} \frac{1}{e^{2xn}(2n+1)^{m+1}}$ . @Lucian : Yes, you are right. And the problem seems to be $\sqrt{.}$ of $\sqrt{e^{2t}-4}$ . I don't know literatur for such types of integrals (right side of the equation chain). ;) – user90369 May 03 '19 at 08:52

0 Answers0