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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 Catalan numbers?

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At least we can compute $a_1$. Notice that

$$ a_1 = \sum_{k=1}^{\infty} \frac{k!(k+1)!}{(2k)!} = \sum_{k=1}^{\infty} k(k+1) \int_{0}^{1} t^{k}(1-t)^{k-1} \,dt = \int_{0}^{1} \frac{2t}{(1-t+t^2)^3} \, dt. $$

This can be computed by applying the substitution $t=\frac{1}{2}+\frac{\sqrt{3}}{2}\tan\theta$ to obtain

$$ a_1 = 1+\frac{4\pi}{9\sqrt{3}}. $$


Addendum. Similar approach leads to a much more complicated expression

$$ a_2 = 1024 \int_{-1}^{1}\int_{-1}^{1} \frac{321 - 66(u^2+v^2) + 68u^2v^2 + u^4 + v^4 - 2u^2v^2(u^2+v^2) + u^4v^4}{(15 + u^2 + v^2 - u^2v^2)^5} \, dudv $$

which I have no idea how to simplify (and is likely to be impossible).

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    $\begingroup$ FWIW, Maple can do one integral: $$ a_2 = -64\,\int_{-1}^{1}\!{\frac {1}{ \left( {v}^{2}+15 \right) ^{4}} \left( -12\,{\frac {{v}^{4}-50\,{v}^{2}+145}{\sqrt {{v}^{4}+14\,{v}^{ 2}-15}}{\rm arctanh} \left({\frac {{v}^{2}-1}{\sqrt {{v}^{4}+14\,{v}^{ 2}-15}}}\right)}+25\,{v}^{2}-465 \right) }\,{\rm d}v$$ $\endgroup$ – Robert Israel May 29 '18 at 22:44

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