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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.

Infiniticism
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3 Answers3

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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)?

Update

$\operatorname{Re} \operatorname{Li}_4 \left (\frac{1 + i}{2} \right )$ 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.

omegadot
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  • @ omegadot Are you sure that we have $H_4(z) = \sum_{n=1}^{\infty} H_{n} \frac{z^n}{n^4}$? And if so, where can I find the expression? Wasn't the answer of Tunk-Fey proved wrong? https://math.stackexchange.com/questions/3366039/a-group-of-important-generating-functions-involving-harmonic-number – Dr. Wolfgang Hintze Apr 09 '20 at 02:35
  • @Dr. Wolfgang Hintze - It is listed under Eq. (4) in his solution. I have not checked if the expression he gives is correct but judging from the comments below his answer it seems it is not. I have updated my solution to reflect this. – omegadot Apr 09 '20 at 04:22
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    @ omegadot you can easyliy falsify the formula of Tunk Fey for $H_4$ by considering his expression in the vicinity of the origin which is a logarithmic divergence proportional to $\log(x)$ whereas is should be analytic starting as $x + \frac{x^2}{2^5} $ – Dr. Wolfgang Hintze Apr 10 '20 at 08:39
  • @Display name - Just found the value for $\operatorname{Re} \operatorname{Li}_4$ in your wonderful paper. Thanks. – omegadot Apr 11 '20 at 00:24
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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 $2007.03957$)

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

Infiniticism
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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.

Dr. Wolfgang Hintze
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  • I always wondered why some non-elementary functions are considered to be more "elementary" than the others. For example why the hypergeometric function is not so "nice" as the polylog or polygamma ones. :) – user Apr 08 '20 at 13:38
  • Isn't it absolutely common in mathematics that from some point of the value of a certain parameter an expression cannot be expressed by a standard set of elementary expression? Think of the roots of polynomials which are not expressable in general by radicals if the degreee exceeds 5. But in this example there is a non-existence proof which, as far as I know, is not availble for the problem of the OP. – Dr. Wolfgang Hintze Apr 08 '20 at 14:17