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?


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).

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.