Evaluate $\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$ in terms of elementary constants

How can we evaluate $$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$$

Related: $\sum _{n=1}^{\infty } \frac{1}{n^4 2^n \binom{3 n}{n}}$ is solved recently (see here for the solution) by elementary method, so I wonder if a generalization to weight $5$ can be made.

Beta+IBP $3$ times+log factorization yields

• $S=\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}=\int_0^1 \frac2x \text{Li}_4\left(\frac{x^2(1-x)}2\right) dx\\ =\int_0^1 -\frac{2\left((3 x-2) \text{Li}_3\left(\frac{1}{2} (1-x) x^2\right)\right) \log (x)}{(x-1) x} dx\\ =\int_0^1 \frac{2\left((3 x-2) \text{Li}_2\left(\frac{1}{2} (1-x) x^2\right)\right) \left(\log ^2(x)-\text{Li}_2(1-x)\right)}{(x-1) x} dx\\ =\int_0^1 \left(\frac{2}{x}-\frac{1}{1-x}\right) f(x) \left(\log \left((1-x)^2+1\right)+\log (x+1)-\log (2)\right) dx$

Where

• $\small f(x)=2\left(-\text{Li}_3(1-x)+2 \text{Li}_3(x)-2 \text{Li}_2(1-x) \log (x)-2 \text{Li}_2(x) \log (x)+\frac{2 \log ^3(x)}{3}-\log (1-x) \log ^2(x)\right)$

Apply reflection one obtain

• $S=\int_0^1 \left(\frac{2 f(1-x) \left(\log \left(x^2+1\right)-\log (2)\right)}{1-x}-\frac{f(1-x) \log \left(x^2+1\right)}{x}+\frac{2 f(x) \log (x+1)}{x}-\frac{f(x) (\log (x+1)-\log (2))}{1-x}\right) \, dx$

Which has $4$-admmisible integrand thus solvable via MZVs of level $4$ (see arXiv $2007.03957$)

$$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}=3 \pi \beta (4)+4 \pi \operatorname{Im} \left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{51 \text{Li}_5\left(\frac{1}{2}\right)}{2}-15 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+\frac{\pi ^2 \zeta (3)}{4}+\frac{9 \zeta (5)}{2}-3 \zeta (3) \log ^2(2)-\frac{97 \log ^5(2)}{240}+\frac{41}{144} \pi ^2 \log ^3(2)-\frac{61}{960} \pi ^4 \log (2)$$

This is by no means a solution but an extended comment which shows that this sum - generalized to a generating function - belongs to the well-known class of special functions, the hypergeometric functions. Maybe this can be of some use.

Defining the generating function

$$g(q,z) = \sum _{n=1}^{\infty } \frac{z^n}{n^q\binom{3 n}{n} }\tag{1}$$

the left hand side of the identity in question is the generating function for $q=5$ at the point $z=\frac{1}{2}$.

$$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}} = g(5,\frac{1}{2})\tag{2}$$

We have

$$g(0,z)= \sum _{n=1}^{\infty } \frac{z^n}{\binom{3 n}{n}}=\frac{z}{3} \; _3F_2\left(1,\frac{3}{2},2;\frac{4}{3},\frac{5}{3};\frac{4 z}{27}\right)\tag{3}$$

and, recursively, with the indefinite integral

$$g(q+1,z)=\int \frac{1}{z} g(q,z)\,dz\tag{4}$$

Explicitly,

$\begin{align} & g(1,z) = \frac{z}{3} \; _3F_2\left(1,1,\frac{3}{2};\frac{4}{3},\frac{5}{3};\frac{4 z}{27}\right)\\\\ & g(2,z) =\frac{z}{3} \; _4F_3\left(1,1,1,\frac{3}{2};\frac{4}{3},\frac{5}{3},2;\frac{4 z}{27}\right)\\\\ &g(3,z)= \frac{z}{3} \; _5F_4\left(1,1,1,1,\frac{3}{2};\frac{4}{3},\frac{5}{3},2,2;\frac{4 z}{27}\right)\\\\ \ldots\\\\ & g(q,z) =\frac{z}{3} \; _P F_Q\left(1_{1},1_{2},1_{3},\ldots,1_{q+1},\frac{3}{2};\frac{4}{3},\frac{5}{3},2_{1},2_{2},\ldots,2_{q-1};\frac{4 z}{27}\right), \\\\ & P=q+2, Q=q+1\\ \end{align}\tag{5}$

In the end we are interested in these functions at $z=\frac{1}{2}$.

Hopefully there are some means of expanding these hypergeometric functions in terms of the more common function types like zeta, polylog, polygamma (specifically harmonic number).

Extension

Similarly we can study the family of sums

$$g_{k,q}(z)=\sum _{n=1}^{\infty }\frac{1}{n^q} \frac{z^n}{\binom{k\; n}{n}}\tag{6}$$

These generating functions are also expressible in terms of hypergeometric functions.