TITLE: How find the value $\sum\limits_{k=1}^{\infty}(C_{3k}^k)^{-1}$
QUESTION [5 upvotes]: find the value 
$$\sum_{k=1}^{\infty}\dfrac{1}{C_{3k}^{k}}$$
and  long ago,I have see this and is easy
$$\sum_{k=1}^{\infty}\dfrac{1}{C_{2k}^{k}}$$
where 
$$C_{n}^{k}=\dfrac{n!}{(n-k)!k!}$$

REPLY [9 votes]: We can write
\begin{align}
S=\sum_{k=1}^{\infty}{3k \choose k}^{-1}&=\sum_{k=1}^{\infty}\frac{\Gamma(k+1)\Gamma(2k+1)}{\Gamma(3k+1)}=\sum_{k=1}^{\infty}kB(k,2k+1)=\\
&=\sum_{k=1}^{\infty}k\int_0^1x^{k-1}(1-x)^{2k}dx=\\
&=\int_{0}^1\frac{(1-x)^2dx}{\left[1-x(1-x)^2\right]^2}=\\
&=\int_{0}^1\frac{x^2dx}{\left(1-x^2+x^3\right)^2}.
\end{align}
The integrand in the last expression is a rational function of $x$. Therefore the antiderivative can be calculated using standard methods in terms of rational functions and logarithms and the result will be expressed in terms of the roots of the cubic equation $x^3-x^2+1=0$.

UPD: Added the actual value:
$$S=\frac{4}{23}-\frac{2}{23}\sum_{k=1}^3\frac{\left(x_k+3\right)\ln(1-x_k)}{3x_k^2-1},$$
where $x_{1,2,3}$ denote three solutions of the cubic equation $x^3-x+1=0$.