It is known that the following functions are elliptic integrals $$ \, _2F_1\left({a,1-a\atop 1};x\right),\quad a=\tfrac12,\tfrac13,\tfrac14,\tfrac16,\tag{1} $$ $$ \, _2F_1\left({\tfrac{1}{3},\tfrac{2}{3}\atop \tfrac{4}{3}};x\right),\, _2F_1\left({\tfrac{1}{2},b\atop b+1};x\right),\quad b=\tfrac14,\tfrac16.\tag{2} $$ It was proved in the arxiv preprint by Martin Nicholson that the generating function $$ \sum_{n=1}^\infty \frac{(a)_n(1-a)_n}{(n!)^2}H_nx^n,\quad a=\tfrac12,\tfrac13,\tfrac14,\tfrac16\tag{3} $$ has closed form in terms of elliptic integrals (1). The formulas for $a=\tfrac12$ can be found in this question. However no similar series with harmonic numbers are known for (2). Now consider the series of Kummer's and Pfaff's transformations of hypergeometric function $$ \, _2F_1\left({\tfrac{1}{2},b\atop b+1};4 x (1-x)\right)=\, _2F_1\left({1,2b\atop b+1};x\right)=\frac{1}{1-x}\, _2F_1\left({1,1-b\atop b+1};\frac{x}{1-x}\right). $$ We see that the series $$ \, _2F_1\left({1,\tfrac34\atop \tfrac54};x\right) $$ is related to elliptic integral (2) with $b=\tfrac14$. It follows from Theorem 4 of the preprint mentioned above that there is the following strange evaluation of a sum with harmonic numbers \begin{equation} \sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n}\frac{\tfrac14-(-1)^n}{2n+1}H_n=\frac{\Gamma^4(\tfrac14)}{64\pi}\ln 2.\tag{4} \end{equation} We see that two series with different 'arguments' $x=1$ and $x=-1$ conspire to have a closed from. However attempts to find closed form for each sum $$ \sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n}\frac{H_n}{2n+1},~\sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n}\frac{(-1)^nH_n}{2n+1}\tag{5} $$ separately, including the method outlined in the preprint, was not successful so far.

Q1: Do the series (5) have closed form?

Q2: Does the series $$ \sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n}\frac{H_n}{2n+1}\left(2(-x)^n-\frac{(4x)^n}{(1+x)^{2n+1}}\right)\tag{6} $$ have closed form for all $~|x|\le 1$?

Of course (6) is not a generating function, but at least the sum with $x=1$ in (6) can be calculated according to (4).

  • $\begingroup$ I do believe we can find closed forms in terms of generalized hypergeometric functions for any of these series because the ratio of consequent terms is a rational function of $n$. $\endgroup$ – Yuriy S Feb 2 '18 at 8:03
  • $\begingroup$ @YuriyS representation in terms of generalized hypergeometric functions will do, so if you have it please feel free to add it as an answer. $\endgroup$ – user82588 Feb 2 '18 at 8:11
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    $\begingroup$ I was definitely mistaken, it's not as simple, because the ratio of harmonic numbers (even while always being a rational function of $n$) has more and more factors as $n$ grows, and this can't be directly converted to hypergeometric form $\endgroup$ – Yuriy S Feb 2 '18 at 9:18
  • $\begingroup$ @YuriyS thanks for your answer and interest in this question. Probably I will accept your answer, because it is unlikely that there is anything more that can be said about this sum. $\endgroup$ – user82588 Feb 3 '18 at 7:58

Let me share some results for the integral form of the series:

$$S(p)=\sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n}\frac{H_n}{2n+1} p^n$$

Where $p$ can be $ \pm 1$ or any other number provided the series converges.

First, we will use the integral expression for $H_n$:

$$H_n=\int_0^1 \frac{1-x^n}{1-x} dx$$

Now we exchange the summation and the integration, keeping in mind that the terms inside the integral diverge separately and we are only allowed to consider them as such under the integral sign.

$$S(p)=\int_0^1 \frac{dx}{1-x} \left( \sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n (2n+1)} p^n-\sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n (2n+1)} (px)^n \right)$$

In other words, we need to find the closed form for:

$$S_1(p)=\sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n (2n+1)} p^n$$

To get this into the hypergeometric form we transform the general term:

$$\frac{(\tfrac34)_n}{(\tfrac54)_n (2n+1)}=\frac{(\tfrac34)_n (1)_n}{2(\tfrac54)_n (n+1/2)n!}=\frac{1}{2} \frac{\Gamma(1/2)}{\Gamma(3/2)} \frac{(\tfrac34)_n (1)_n (\tfrac12)_n}{(\tfrac54)_n (\tfrac32)_n n!}$$

In other words:

$$S_1(p)={_3 F_2} \left( 1, \frac{1}{2}, \frac{3}{4}; \frac{3}{2}, \frac{5}{4}; p \right)-1$$

Which means:

$$S(p)=\int_0^1 \frac{dx}{1-x} \left( {_3 F_2} \left( 1, \frac{1}{2}, \frac{3}{4}; \frac{3}{2}, \frac{5}{4}; p \right)-{_3 F_2} \left( 1, \frac{1}{2}, \frac{3}{4}; \frac{3}{2}, \frac{5}{4}; px \right) \right)$$

To simplify this further and get rid of special functions, let's use the reduction formula (see Wikipedia):

$${_3 F_2} \left( a_1, a_2, a_3; b_1, b_2; p \right)=\frac{\Gamma(b_2)}{\Gamma(a_3) \Gamma(b_2-a_3)} \int_0^1 t^{a_3-1} (1-t)^{b_2-a_3-1} {_2 F_1} \left( a_1, a_2; b_1; p t \right) dt$$

We are free to choose which parameters to reduce, so let's get rid of the most complicated ones:

$${_3 F_2} \left( 1, \frac{1}{2}, \frac{3}{4}; \frac{3}{2}, \frac{5}{4}; p \right)=\frac{\Gamma(5/4)}{\Gamma(3/4) \Gamma(1/2)} \int_0^1 t^{-1/4} (1-t)^{-1/2} {_2 F_1} \left( 1, \frac{1}{2}; \frac{3}{2}; p t \right) dt$$

And finally, the wonderful thing is:

$${_2 F_1} \left( 1, \frac{1}{2}; \frac{3}{2}; p t \right)=\frac{\tanh^{-1} \sqrt{p t} }{\sqrt{p t}}$$


$$S(p)=\frac{\Gamma(5/4)}{\Gamma(3/4) \sqrt{ \pi}} \int_0^1 \int_0^1 \frac{dx dt}{(1-x)t^{1/4} \sqrt{1-t}} \left( \frac{\tanh^{-1} \sqrt{p t} }{\sqrt{p t}} - \frac{\tanh^{-1} \sqrt{p x t} }{\sqrt{p x t}} \right) $$

We obtain the integral form valid for all the sums (5) and (6).

In addition, the arctangents also have a simple integral form, which makes this function expressible as a triple integral of algebraic functions.

$$\frac{\tanh^{-1} \sqrt{p t} }{\sqrt{p t}}=\int_0^1 \frac{dy}{1-pty^2}$$

After some simple transformation we get the following triple integral formula over a unit cube:

$$S(p)=\frac{\Gamma(5/4)}{\Gamma(3/4) \sqrt{ \pi}} p \int_0^1 \int_0^1 \int_0^1 \frac{t^{3/4} y^2 ~ dx dt dy}{\sqrt{1-t}(1-p t y^2)(1-p x t y^2)} $$

This result checks out numerically, for example:

enter image description here

Integrating w.r.t. $x$ we get another double integral representation:

$$S(p)=-\frac{\Gamma(5/4)}{\Gamma(3/4) \sqrt{ \pi}} \int_0^1 \int_0^1 \frac{ \log (1-p t y^2) dt dy}{t^{1/4} \sqrt{1-t}(1-p t y^2)} $$

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