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

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!

Answer 1

Without recognizing a binomial transform, you may notice that

$$\begin{eqnarray*}\sum_{n=0}^{k}\binom{k}{n}\frac{1}{2(n+1)(3n+1)} &=& \frac{3}{4}\int_{0}^{1}\sum_{n=0}^{k}\binom{k}{n}\left(x^{3n}-x^{3n+2}\right)\,dx\\&=&\frac{3}{4}\int_{0}^{1}(1-x^2)(1+x^3)^k\,dx\end{eqnarray*}$$ hence: $$\sum_{k\geq 0}\frac{1}{2^{k+1}}\sum_{n=0}^{k}\binom{k}{n}\frac{1}{2(n+1)(3n+1)}=\frac{3}{4}\int_{0}^{1}\frac{1-x^2}{1-x^3}\,dx $$ and the last integral can be computed through partial fraction decomposition.

Answer 2

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

$$\sum_{n=0}^\infty\frac1{2(n+1)(3n+1)}=\frac1{12}(\pi\sqrt3+9\ln(3))$$