How find the value $\sum\limits_{k=1}^{\infty}(C_{3k}^k)^{-1}$ 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 
$$C_{n}^{k}=\dfrac{n!}{(n-k)!k!}$$
 A: 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 $x^3-x+1=0$.
A: For #2:
$$\sum_{k=1}^{\infty}\dfrac{1}{C_{2k}^{k}}=\frac{1}{3}+\frac{2\pi}{9\sqrt{3}}$$
Because,
$$\sum_{n=0}^\infty\frac{x^n}{C^n_{2n}}=\frac{4(\sqrt{4-x}+\sqrt{x}\arcsin(\frac{\sqrt{x}}{2}))}{\sqrt{(4-x)^3}}$$
Which is because
$$\text{If we set, } A_n=\frac{1}{C^{n}_{2n}}$$
Then
$$(4n+2)A_{n+1}=(n+1)A_n$$
And so,
$$\sum_{n=0}^\infty(4n+2)A_{n+1}x^n=\sum_{n=0}^\infty(n+1)A_nx^n$$
So that if $$A(x)=\sum_{n=0}^\infty A_nx^n$$
$$\frac{d}{dx}A(x)-\frac{x+2}{x(4-x)}A(x)=\frac{-2}{4x-x^2}$$
Now with some work you can verify $A(x)$ satisfies the above differential equation.
For #1:
Likewise 
If you let $K_n=\frac{1}{C^n_{3n}}$
And,
$$F(x)=\sum_{n=1}^\infty\frac{x^n}{C^n_{3n}}=\sum_{n=1}^\infty K_nx^n$$
We have that, $3(3n+1)(3n+2)K_{n+1}=2(n+1)(2n+1)K_{n}$
So,
$$\sum_{n=1}^\infty 3(3n+1)(3n+2)K_{n+1}x^n =\sum_{n=1}^\infty 2(n+1)(2n+1)K_{n}x^n$$
And so that we get,
$$4x^2\frac{d^2}{dx^2}F(x)-17x\frac{d}{dx}F(x)-\frac{4}{1-x}=0$$
Which is a second order linear ordinary differential equation, and can be solved in terms of elementary functions, namely in terms of logarithms and inverse tangent functions.
Once you have $F$, then just compute $$F(1)=\sum_{n=1}^\infty\frac{1}{C^n_{3n}}$$
