Bino Harmonic Series 2

The complete elliptic function of the first kind (under one convention) is $K(k) \; = \; \int_0^{\frac12\pi} \frac{d\theta}{\sqrt{1-k^2\sin^2\theta}} \; = \; \tfrac12\pi \sum_{n=0}^\infty {2n \choose n}^2 \frac{k^{2n}}{2^{4n}}$ so that $\sum_{n=0}^\infty {2n \choose n}^2 \frac{k^n}{2^{5n}} \; = \; \tfrac{2}{\pi}K\left(\sqrt{\tfrac{k}{2}}\right)$ Thus, if we write $S \; = \; \sum_{n=1}^\infty {2n \choose n}^2 \frac{H_n}{2^{5n}}$ then $\begin{array}{rcl} \displaystyle S & = & \displaystyle \sum_{n=1}^\infty {2n \choose n}^2 2^{-5n} \int_0^1 \frac{1-k^n}{1-k}\,dk \; = \; \frac{2}{\pi}\int_0^1 \frac{K\left(\sqrt{\frac12}\right) - K\left(\sqrt{\frac{k}{2}}\right)}{1-k}\,dk \\ & = & \displaystyle \frac{2}{\pi} \int_0^1 \frac{dk}{1-k}\int_0^{\frac12\pi} \left\{ \frac{1}{\sqrt{1 - \frac12\sin^2\theta}} - \frac{1}{\sqrt{1 - \frac12k\sin^2\theta}}\right\}\,d\theta \\ & = & \displaystyle \frac{1}{\pi}\int_0^1\,dk \int_0^{\frac12\pi} \frac{\sin^2\theta\,d\theta}{\sqrt{(1-\frac12\sin^2\theta)(1 - \frac12k\sin^2\theta)}\left(\sqrt{1 - \frac12\sin^2\theta} + \sqrt{1 - \frac12k\sin^2\theta}\right)} \\ & = & \displaystyle \frac{1}{\pi} \int_0^{\frac12\pi} \frac{\sin^2\theta\,d\theta}{\sqrt{1-\frac12\sin^2\theta}} \left(\int_0^1 \frac{dk}{\sqrt{1 - \frac12k\sin^2\theta}\left(\sqrt{1-\frac12\sin^2\theta} + \sqrt{1-\frac12k\sin^2\theta}\right)}\right) \\ & = & \displaystyle \frac{2}{\pi} \int_0^{\frac12\pi} \frac{d\theta}{\sqrt{1-\frac12\sin^2\theta}}\left\{ 2\ln\left(1 + \sqrt{1 - \tfrac12\sin^2\theta}\right) - 2\ln2 - \ln\left(1 -\tfrac12\sin^2\theta\right)\right\} \end{array}$ Since $\tfrac{1}{\sqrt{2}}$ is equal to its own complement, we have (Gradshteyn and Rhyzik) $\int_0^{\frac12\pi} \frac{\ln\left(1 + \sqrt{1 - \tfrac12\sin^2\theta}\right)}{\sqrt{1 - \frac12\sin^2\theta}}\,d\theta \; = \; \tfrac14(\pi - \ln2)K\big(\tfrac{1}{\sqrt{2}}\big)$ while this paper tells us that $\int_0^{\frac12\pi} \frac{\ln\left(1 - \frac12\sin^2\theta\right)}{\sqrt{1 - \frac12\sin^2\theta}}\,d\theta \; = \; -\tfrac12\ln2 K\big(\tfrac{1}{\sqrt{2}}\big)$ (on further looking, this integral is in G&R as well) and hence $\begin{array}{rcl} \displaystyle S & = & \displaystyle \tfrac{2}{\pi}\left[ 2\times\tfrac14(\pi - \ln2)K\big(\tfrac{1}{\sqrt{2}}\big) - 2\ln2 K\big(\tfrac{1}{\sqrt{2}}\big) + \tfrac12\ln2 K\big(\tfrac{1}{\sqrt{2}}\big) \right] \\ & = & \displaystyle \tfrac{1}{\pi}K\big(\tfrac{1}{\sqrt{2}}\big)\big(\pi - 4\ln2\big) \; = \; \frac{\sqrt{\pi}}{2\Gamma\left(\frac34\right)^2}\big(\pi - 4\ln2\big) \end{array}$ making the answer $1+2+3+4+1+4+2 = \boxed{17}$ .