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


Beta+IBP $3$ times+log factorization yields

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


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

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

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


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

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).


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.

This is not an answer, but an approach to how one could find your sum in principle.

As I see it, it all comes down to finding a generating function for $$\sum_{n = 1}^\infty \frac{H_n x^n}{n^4}. \tag1$$ Once found, substituting $x = (1 + i)/2$ into it and taking the real part should deliver the result.

Can this generating function actually be found I do not know. One attempt (which is incorrect) to find it can be found here under Eq. (4). I expect such an expression to contain polylogarithmic terms up to order five. I am not sure $$\operatorname{Re} \operatorname{Li}_4 \left (\frac{1 + i}{2} \right ),$$ is reducible to more fundamental constants like $\pi$, $\zeta (3)$, $\log (2)$, $\operatorname{Li}_4 \left (\frac{1}{2} \right )$, let alone $$\operatorname{Re} \operatorname{Li}_5 \left (\frac{1 + i}{2} \right ).$$

So it boils down to, is it possible to find the generating function for (1)?


$\operatorname{Re} \operatorname{Li}_4 \left (\frac{1 + i}{2} \right )$ is reducible to more fundamental constants. Here \begin{align} \operatorname{Re} \operatorname{Li}_4 \left (\frac{1 + i}{2} \right ) &= \frac{5}{16} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{343}{1024} \zeta (4) - \frac{5}{128} \zeta (2) \log^2 (2) + \frac{1}{96} \log^4 (2)\\ &=\frac{5}{16} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{343 \pi^4}{92160} - \frac{5 \pi^2}{768} \log^2 (2) + \frac{1}{96} \log^4 (2). \end{align} This value can be found in the paper given here.