We will be considering series such as this :
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{1/6+r}{2^{8r}}\right)H_{r-1/2}=-\frac{8\ln2}{9\pi}$$
Consider $K(k)$ and $E(k)$ to be the Complete Elliptical Integrals of the First and Second Kind Respectively with $K'(k)=K(k')$ and $E'(k)=E(k')$, where $k^2+k'^2=1$.
First let us consider the following Generating Functions:
$$\sum_{r=0}^{\infty}\binom{2r}{r}^2\frac{H_r}{2^{4r}}k^{2r}=K'(k)+\frac{2K(k)}{\pi}\ln\left(\frac{k}{4k'}\right)$$
$$\sum_{r=0}^{\infty}\binom{2r}{r}^2\frac{H_{2r}}{2^{4r}}k^{2r}=\frac{K'(k)}{2}+\frac{2K(k)}{\pi}\ln\left(\frac{\sqrt{k}}{2k'}\right)$$
Now one may let $k\to k\sin t$ and Integrate from $0$ to $\pi/2$ with respect to $t$.
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{H_r}{2^{6r}}\right)(2kk')^{2r}=\frac{4K(k)K'(k)}{3\pi}+\frac{8K^2(k)}{3\pi^2}\ln(kk'/4)-\frac{2v(2kk')}{\pi^2}$$
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{H_{2r}}{2^{6r}}\right)(2kk')^{2r}=\frac{2K(k)K'(k)}{3\pi}+\frac{4K^2(k)}{3\pi^2}\ln(kk'/4)-\frac{2v(2kk')}{\pi^2}$$ where, $$v(k)=\int_{0}^{\pi/2}K(k\sin t)\ln(1-k^{2}\sin^{2}t)dt$$
What's troublesome here is the function $v(k)$, I am not able reduce it further.
At the very least we can find the Generating Function for $H_{n}-H_{2n}=\overline{H}_{2n}$ which is the Skew Harmonic Number (As the Function $v(k)$ cancels out).
$$\pi\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{\overline{H}_{2r}}{2^{6r}}\right)(2kk')^{2r}=\frac{2}{3}K(k)K'(k)+\frac{4}{3\pi}\ln(kk'/4)K^2(k)$$
Now we differentiate here and write everything in Terms of $k_n$, $\alpha_n,\beta_n$.
Where $k_n$ is the Elliptic Integral Singular Value such that $K'/K=\sqrt{n}$. $\alpha_n$ is the Elliptic Alpha Function.
and $\beta_n$ is defined by us as $$\beta_n=\ln(k_nk'_n/4)$$
Now,
$$\pi\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{\overline{H}_{2r}}{2^{6r}}\right)(2k_nk'_n)^{2r}=\frac{2}{3}K_n^2\left(\sqrt{n}+\frac{2\beta_n}{\pi}\right)$$
and, $$\pi\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{\overline{H}_{2r}}{2^{6r}}\right)r(2k_nk'_n)^{2r}=\frac{2}{3}K_n^2\left(\frac{1}{\pi}+\left(\frac{\alpha_n-k_n^2\sqrt{n}}{2k_n^2-1}\right)\left(1+\frac{2\beta_n}{\pi\sqrt{n}}\right)\right)-\frac{\beta_n}{3(2k_n^2-1)\sqrt{n}}$$
Although we cannot Eliminate $K_n$ by taking a Linear Combination, we can remove the Algebraic Term attached with $K_n$.
This gives us:
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{\overline{H}_{2r}}{2^{6r}}\right)(2k_nk'_n)^{2r}\left(\frac{k_n^{2}-\alpha_n/\sqrt{n}}{k_n^2-k'^2_n}+r\right)=\frac{1}{\pi}\left(\frac{2K^2_n}{3\pi}-\frac{\beta_n}{3\sqrt{n}(k_n^2-k_n'^2)}\right)$$
Some Particular Evaluations of these using $k_3k'_3=1/4$, $k_7k'_7=1/16$ are:
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{\color{blue}{1/6}+r}{2^{8r}}\right)\overline{H}_{2r}=\frac{2}{3\pi}\left(\frac{K^2(k_{3})}{\pi}\ -\frac{4\ln2}{3}\right)$$
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{\color{blue}{5/42}+r}{2^{12r}}\right)\overline{H}_{2r}=\frac{8}{21\pi}\left(\frac{7K^2(k_{7})}{4\pi}-2\ln2\right)$$
And recall Ramanujan's Famous Series for $\pi$: $$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{\color{blue}{1/6}+r}{2^{8r}}\right)=\frac{2}{3\pi}$$
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{\color{blue}{5/42}+r}{2^{12r}}\right)=\frac{8}{21\pi}$$
Which is Interesting, I am guessing the Algebraic Term along with $r$ also appears in the Derivation of these Series.
This got me thinking that although we do not have the closed form of Generating Functions for $H_r$ and $H_{2r}$ (due to function $v(x)$) so instead I set up to use Integer Relation Algorithms which surprisingly gives.
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{1/6+r}{2^{8r}}\right)H_{r}=\frac{4}{3\pi}\left(\frac{K^2(k_{3})}{\pi}\ -\ln2\right)$$
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{1/6+r}{2^{8r}}\right)H_{2r}=\frac{2}{9\pi}\left(\frac{3K^2(k_{3})}{\pi}\ -2\ln2\right)$$
Nothing for the case of $k_7$. (If it involves roots then I don't know how to use Integer Relations).
This means we have a certain Hope of Evaluating $v(k)$. Both of the above equations are analogous to: $$2v\left(\frac{1}{2}\right)+3v'\left(\frac{1}{2}\right)=-\frac{8\pi \ln2}{3}\tag{1}$$
In fact now we can take a Linear Combination to Eliminate $K^2(k_3)$ as follows:
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{1/6+r}{2^{8r}}\right)(2H_{2r}-H_r)=\frac{4\ln2}{9\pi}$$
Which is a good start.
If one wants something fancier then use $H_{n-1/2}=2H_{2n}-H_{n}-2\ln2$ to get:
$$\boxed{\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{1/6+r}{2^{8r}}\right)H_{r-1/2}=-\frac{8\ln2}{9\pi}}$$
Questions:
- Do we know anything about the Function $v(k)$? The question about Generating Functions for Cubed Binomial $H_{r}$ and $H_{2r}$ has been considered in this post Generating function of the sequence $\binom{2n}{n}^3H_n$ but no response so far.
- Can we prove $(1)$?
- Like the Higher Order Variations of the $1/\pi^m$ Series, are higher order variations of the Binomial Harmonic Sums been considered anywhere in Literature?
Not an answer.
Only to give you a related formula. It can be proven also. $$\frac{2}{\pi}K(k)=4\sum_{n=1}^{\infty}\binom{2n}{n}^2\frac{n^3k^{2n-2}H_{n}}{2^{4n}(2n-1)^2}-\sum_{n=1}^{\infty}\binom{2n}{n}^2\frac{(n+1)k^{2n}H_{n}}{2^{4n}}\tag{1}.$$ Multiplying both sides of $(1)$ by $\frac{1}{\sqrt{1-k^2}} $ and integrating in $(0,1)$ gives the very slow convergent series: $$\sum_{n=1}^{\infty}\binom{2n}{n}^3\frac{(4n^3+6n^2-5n+1)H_{n}}{2^{6n}(2n-1)^3}=\frac{\pi}{\Gamma(3/4)^4}\tag{2}.$$ Others with better convergence can be obtained: $$\sum_{n=1}^{\infty}\binom{2n}{n}^3\frac{(24n^4+4n^3+6n^2-5n+1)H_{n}}{2^{8n}(2n-1)^3}=\frac{108^{1/6}\Gamma{(1/3)}^6}{8\pi^4}\tag{3}.$$ $$\sum_{n=1}^{\infty}\binom{2n}{n}^3\frac{(72n^4-4n^3-6n^2+5n-1)(-1)^{n}H_{n}}{2^{9n}(1-2n)^3}=\frac{32\sqrt{2}\Gamma{(5/4)}^4}{\pi^3}\tag{4}.$$ $$\sum_{n=1}^{\infty}\binom{2n}{n}^3\frac{(504n^4+4n^3+6n^2-5n+1)H_{n}}{2^{12n}(2n-1)^3}=\frac{\Gamma{(1/7)}^2\Gamma{(2/7)}^2\Gamma{(4/7)}^2}{4\sqrt{7}\pi^{4}}\tag{5}.$$ As you see with $(1)$ seems that $\log{2}$ disappears and maybe you will arrive to series for $1/\pi$ involving also harmonic coefficients. Seeing the relation pattern with Ramanujan's series for the theory $s=1/2$ seems that we have associated series involving also harmonic numbers in the general terms.
As we said there is a counter part for Ramanujan's series for $1/\pi$, here is one example: $$\sum_{n=1}^{\infty}\binom{2n}{n}^3\frac{p(n)(-1)^{n+1}H_{n}}{2^{9n}(2n-1)^3}=\frac{2\sqrt{2}}{\pi}\tag{6},$$ where $p(n)=432n^5-336n^4-40n^3+24n^2-n-1$.