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I am interested in evaluating the sums

$$ \sum_{k=1}^{\infty}{\binom {2k}{k}^{-n}}, $$

where $n$ is a positive integer. It is already known that for $n=1$ we have

$$ \sum_{k=1}^{\infty}\frac{1}{\binom {2k}{k}}=\frac{9+2\sqrt 3 \pi}{27}. $$

Several papers analyze the properties of sums involving the above identity (such as this one), however I was not able to find any material relating to the cases $n>1$. I already know these sums converge for all positive integers $n>0$, however I would be interested in finding a nice closed form for them as in the case $n=1$. How would I go about this? Are there results available about such sums in literature?

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    $\begingroup$ In the $n=2$ case we may try to adapt the technique used for proving the mentioned identity: $$ \sum_{k\geq 1}\binom{2k}{k}^{-2} = \sum_{k\geq 1}\frac{\Gamma(k+1)^4}{\Gamma(2k+1)^2}=\sum_{k\geq 1}k^2 B(k,k+1) B(k,k+1)$$ by the integral representation for Euler's Beta function equals $$ \iint_{(0,1)^2}\frac{(1-x)(1-y)(1+x y-x^2 y-x y^2+x^2 y^2)}{(1-x y+x^2 y+x y^2-x^2 y^2)^3}\,dx\,dy $$ or$$\iint_{(0,\pi/2)^2} \frac{16\sin(x)\sin(y)\left(256+96\sin^2(x)\sin^2(y)+\sin^4(x)\sin^4(y)\right)}{\left(16-\sin^2(x)\sin^2(y)\right)^3}\,dx\,dy$$ which can be probably managed through polylogarithms. $\endgroup$
    – Jack D'Aurizio
    Jun 27, 2018 at 13:52
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    $\begingroup$ The evaluation of the previous integral reduces to the evaluation of $$ \int_{0}^{1}\frac{\arcsin(y/4)}{\sqrt{(1-y^2)(16-y^2)}}\,dy$$ or the evaluation of $$\int_{0}^{1/4}\int_{0}^{1}\frac{du}{\sqrt{(1-u)(16-u)(1-a u)}}\,da=\int_{0}^{1}\frac{du}{\sqrt{(1-u)(16-u)}(2+\sqrt{4-u})},$$ by Feynman's trick. $\endgroup$
    – Jack D'Aurizio
    Jun 27, 2018 at 14:09
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    $\begingroup$ @JackD'Aurizio Wow, that's amazing! I wish I could award +1 multiple times $\endgroup$
    – Klangen
    Jun 27, 2018 at 14:11

1 Answer 1

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I think that, for $n>1$, you are entering in the world of hypergeometric functions.

Just have a look to the table and notice the patterns $$\left( \begin{array}{cc} n & S_n \\ 2 & \frac{1}{4} \, _3F_2\left(1,2,2;\frac{3}{2},\frac{3}{2};\frac{1}{16}\right) \\ 3 & \frac{1}{8} \, _4F_3\left(1,2,2,2;\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{64}\right) \\ 4 & \frac{1}{16} \, _5F_4\left(1,2,2,2,2;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{25 6}\right) \\ 5 & \frac{1}{32} \, _6F_5\left(1,2,2,2,2,2;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{ 2};\frac{1}{1024}\right) \end{array} \right)$$ What is interesting is that $\log(S_n)$ is almost a linear function of $n$ (almost $\log(S_n)=-n \log(2)$).

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  • $\begingroup$ Thank you for your insightful notes. Shortly after you wrote it, I tried FunctionExpanding it with Mathematica, and I got: ${\sum_{k=1}^{\infty}{\binom {2k}{k}^{-n}} } = \pi ^{n/2} \sum _k^{\infty } \left(\frac{2^{2 k} \Gamma \left(k+\frac{1}{2}\right)}{\Gamma (k+1)}\right)^{-n}$. How can I prove this? $\endgroup$
    – Klangen
    Jun 27, 2018 at 11:33
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    $\begingroup$ @preservedfruit: You prove this by rewriting the binomial coefficient with the gamma function and use the duplication formula for $\Gamma(2k+1)$. $\endgroup$
    – gammatester
    Jun 27, 2018 at 13:36

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