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I posted the question https://math.stackexchange.com/questions/2799068/is-the-value-of-sum-limits-k-1%e2%88%9e-frac1c-kn-known before on mathstackexchange but realised that it might be more appropriate for mathoverflow after seeing the answer.

Is the value of the sum $a_n:=\sum\limits_{k=1}^{\infty}\dfrac{1}{(C_k)^n}$ known for $n \geq 1$, where $C_k= \dfrac{1}{k+1} \dbinom{2k}{k}$ are the Catalan numbers?

In the mathstackexchange thread it was shown that $a_1= 1+\frac{4\pi}{9\sqrt{3}}$ and that calculating $a_2$ might be more complicated.

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    $\begingroup$ I think the following is plausible. GIven a sequence $c_0, c_1,\dots$ of complex numbers for which the sum $\sum c_k$ is absolutely convergent, the sum $\sum_{k\geq 0}c_k^n$ has a closed form expression (for any reasonable definition of "closed form") for all integers $n\geq 1$ if and only if $\sum c_kx^k$ is a rational function. $\endgroup$
    – Richard Stanley
    Commented Jun 3, 2018 at 23:04
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    $\begingroup$ What about $c_k = \frac{1}{(k+1)^2}$? $\endgroup$
    – Achim Krause
    Commented Jun 4, 2018 at 8:36
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    $\begingroup$ @AchimKrause You are right, my statement needs to be modified. $\endgroup$
    – Richard Stanley
    Commented Jun 4, 2018 at 17:57

2 Answers 2

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It can be written using a hypergeometric function

$${\mbox{$_{n+1}$F$_n$}\left(1,3,\ldots,3;\,\frac32,\ldots,\frac32;\,{\frac{1}{4^n}}\right)}$$

I don't know if further simplification is possible.

EDIT: As requested, here is some elaboration. By definition, this generalized hypergeometric function $f$ is

$$ \sum_{k=0}^\infty \frac{(1)_k \left((3)_k\right)^n}{k!\; \left((3/2)_k\right)^n} (1/4)^{nk}$$ using the pochhammer symbols $$(z)_k = \Gamma(z+k)/\Gamma(z)$$ Thus $(1)_k = k!$, $(3)_k = (k+2)!/2$, $$(3/2)_k =\frac{\Gamma(3/2+k)}{\Gamma(3/2)} = \prod_{j=1}^{k} \left(\frac{2j+1}{2}\right) = \frac{(2k+1)!}{k!\; 4^k}$$ and

$$ f = \sum_{k=0}^\infty \left(\frac{(k+2)! k! }{2\;(2k+1)!}\right)^n = \sum_{k=0}^\infty \frac{1}{(C_{k+1})^n}$$

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    $\begingroup$ Can you elaborate your answer? $\endgroup$
    – user64494
    Commented Jun 3, 2018 at 7:08
  • $\begingroup$ Use the definition of the generalized hypergeometric function in terms of pochhammer symbols, convert to factorials, simplify using the equation $$\dfrac{\Gamma(3/2+k)}{\Gamma(3/2)} = \frac{(2k+1)!}{k!\; 4^k}$$ $\endgroup$
    – Robert Israel
    Commented Jun 3, 2018 at 18:21
  • $\begingroup$ Sorry, your comment is not clear to me though I am PhD for ages. Please elaborate your answer. $\endgroup$
    – user64494
    Commented Jun 3, 2018 at 18:49
  • $\begingroup$ What part of it don't you understand? $\endgroup$
    – Robert Israel
    Commented Jun 3, 2018 at 18:56
  • $\begingroup$ Please start from the convertion of the hypergeometric function to factorials, extending your answer. TIA. $\endgroup$
    – user64494
    Commented Jun 3, 2018 at 19:04
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Let $f(z)=\sum_{n=0}^{\infty}\frac{z^n}{C_n^2}$ is a generating function then it satisfies the differential equation $$z^2(z-16)f''(z)+5z^2f'(z)+4(z-1)f(z)+4=0$$ with initial conditions $f(0)=1$ and $f'(0)=1$. Solving this leads to $$a_2=f(1).$$

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  • $\begingroup$ Can you elaborate your answer? According Maple, the general solution of $z^2(z-16)f''(z)+5z^2f'(z)+4(z-1)+4=0$ is $$f \left( z \right) =-4096\, \left( z-16 \right) ^{-3}-{\frac {409600}{ 3\, \left( z-16 \right) ^{4}}}+128\, \left( z-16 \right) ^{-2}-16\, \left( 3\,z-48 \right) ^{-1}-1/4\,{\frac {{\it \_C1}}{ \left( z-16 \right) ^{4}}}-{\frac {\ln \left( z \right) z \left( {z}^{3}-64\,{z} ^{2}+1536\,z-16384 \right) }{ \left( z-16 \right) ^{4}}}+{\it \_C2} . $$ The function is singular at $z=0$. $\endgroup$
    – user64494
    Commented Jun 3, 2018 at 6:56
  • $\begingroup$ Also up to Maple (the code on demand), the right limit of $f'(z)$ at $z=0$ is $-\infty$. $\endgroup$
    – user64494
    Commented Jun 3, 2018 at 7:13
  • $\begingroup$ @user64494: I am so sorry, I missed the term $f(z)$ from $4(z-1)f(z)$. $\endgroup$
    – T. Amdeberhan
    Commented Jun 3, 2018 at 13:43
  • $\begingroup$ Can you base your modified answer? $\endgroup$
    – user64494
    Commented Jun 3, 2018 at 14:53
  • $\begingroup$ What do you? I don't understand. $\endgroup$
    – T. Amdeberhan
    Commented Jun 3, 2018 at 15:21

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