Simplifying $\sum\limits_{k=0}^{\infty}\frac 1{2^{k+1}}\sum\limits_{n=0}^{\infty}\binom kn\frac 1{2(n+1)(3n+1)}$ 
Question: Is there a way to simplify $$\sum\limits_{k=0}^{\infty}\dfrac 1{2^{k+1}}\sum\limits_{n=0}^{\infty}\binom kn\dfrac 1{2(n+1)(3n+1)}\tag{1}$$
  Into a single summation symbol? $\sum\limits_{k=0}^{\infty}\text{something}$

I inputed it into WolframAlpha and got a really complicated expression$$\sum\limits_{k=0}^{\infty}\dfrac {1-2^{k+1}+3\left(_2F_1\left[\begin{array}{c c}\frac 13,-k\\\frac 43\end{array};-1\right]\right)+3k\left(_2F_1\left[\begin{array}{c c}\frac 13,-k\\\frac 43\end{array};-1\right]\right)}{2^{k+3}(k+1)}$$
Which isn't what I really wanted because the inner sum is significantly more complex than before. Is there a way?
I'm still relatively new to this. If you have a hint, it would mean a lot if you commented it!

Notice that
$$\binom kn=0\forall n>k$$
Thus, it simplifies down to
$$\sum_{k=0}^\infty\frac1{2^{k+1}}\sum_{n=0}^k\binom kn\frac1{2(n+1)(3n+1)}$$
And by an inverse Euler sum, this reduces down to
$$\sum_{n=0}^\infty\frac1{2(n+1)(3n+1)}=\frac1{12}(\pi\sqrt3+9\ln(3))$$
