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 
$$C_{n}^{k}=\dfrac{n!}{(n-k)!k!}$$

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 $$F(1)=\sum_{n=1}^\infty\frac{1}{C^n_{3n}}$$
