Prove $\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$ 
Is there an elegant proof for this identity for all real $s \neq -1, -1/2$?
$$\sum_{n=0}^\infty (-1)^n \binom{s}{n}  \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$$

Where $H_x$ represents harmonic numbers.
I found it in a very roundabout way for $s=0,1,2,3,\dots$, (see this answer and this answer) and I don't actually have the proof for the other identity I used:
$$\sum_{n=0}^\infty \sum_{k=1}^{n+1}\frac{H_{k-\frac{1}{2}}}{k} x^n= \frac{2}{x(1-x)} \left( \operatorname{arctanh}^2 \sqrt{x}+\log 2 \log (1-x) \right)$$
So whichever one you can prove works fine for me, but I think the first series should be easier to prove.
Using Ali Shather's comment, we can simplify:

$$\sum_{n=0}^\infty (-1)^n \binom{s}{n}  \frac{2H_{2n}-H_n}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$$

One of the interesting consequences:
$$\frac{1}{x} \arcsin^2 \sqrt{x}=\sum_{l=0}^\infty \frac{(4 x)^l}{(2l+1) (l+1) {2l \choose l}}$$
 A: Partial solution to the second double sum:
Lets work on the inner sum first
\begin{align}
S_n=\sum_{k=1}^{n+1}\frac{H_{k-\frac12}}{k}=\sum_{k=1}^{n}\frac{H_{k-\frac12}}{k}+\frac{H_{n+\frac12}}{n+1}
\end{align}
by substituting  $\ H_{k-\frac12}=2H_{2k}-H_k-2\ln2$, to have
\begin{align}
\sum_{k=1}^{n}\frac{H_{k-\frac12}}{k}&=2\sum_{k=1}^{n}\frac{H_{2k}}{k}-\sum_{k=1}^{n}\frac{H_k}{k}-2\ln2\sum_{k=1}^{n}\frac{1}{k}\\
&=2\sum_{k=1}^{n}\frac{H_{2k}}{k}-\left(\frac{H_n^2+H_n^{(2)}}{2}\right)-2\ln2H_n\\
\end{align}
Therefore
$$S_n=2\sum_{k=1}^{n}\frac{H_{2k}}{k}-\left(\frac{H_n^2+H_n^{(2)}}{2}\right)-2\ln2H_n+\frac{H_{n+\frac12}}{n+1}\tag{1}$$
Which follows that
$$\sum_{n=0}^\infty\left(\sum_{k=1}^{n+1}\frac{H_{k-\frac12}}{k}\right)x^n=2\underbrace{\sum_{n=0}^\infty\sum_{k=1}^{n}\frac{H_{2k}}{k}x^n}_{\Large S_1}-\underbrace{\sum_{n=0}^\infty\left(H_n^2+H_n^{(2)}\right)x^n}_{\Large S_2}-2\ln2\underbrace{\sum_{n=0}^\infty H_nx^n}_{\Large S_3}+\underbrace{\sum_{n=0}^\infty\frac{H_{n+\frac12}}{n+1}x^n}_{\Large S_4}$$.

Starting with $S_2$ and by using the following identity: ( proved by SuperAbound here)
$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=0}^\infty \left(H_n^2-H_n^{(2)}\right)x^n\tag{2}$$
Add $\ \displaystyle2\sum_{n=0}^\infty H_n^{(2)}x^n=\frac{2\operatorname{Li}_2(x)}{1-x}$ to both sides of $(2)$, we get
$$\boxed{S_2=\frac{2\operatorname{Li}_2(x)+\ln^2(1-x)}{1-x}}$$

$$\boxed{S_3=-\frac{\ln(1-x)}{1-x}}$$

and the last sum
\begin{align}
S_4&=\sum_{n=0}^\infty\frac{H_{n+\frac12}}{n+1}x^n\\
&=\sum_{n=1}^\infty\frac{H_{n-\frac12}}{n}x^{n-1}\\
&=\frac1x\sum_{n=1}^\infty \left(\frac{2H_{2n}-H_n-2\ln2}{n}\right)x^n\\
&=\frac2x\sum_{n=1}^\infty \frac{H_{2n}}{n}x^n-\frac1x\sum_{n=1}^\infty \frac{H_{n}}{n}x^n-\frac{2\ln2}x\sum_{n=1}^\infty \frac{x^n}{n}\\
&=\frac2x\sum_{n=1}^\infty \frac{H_{2n}}{n}x^n-\frac1x(\operatorname{Li}_2(x)+\frac12\ln^2(1-x))-\frac{2\ln2}x(-\ln(1-x))\\
\end{align}
The remaining sum can be simplified as follows:
\begin{align}
\sum_{n=1}^\infty \frac{H_{2n}}{n}x^n&=2\sum_{n=1}^\infty \frac{H_{2n}}{2n}(\sqrt{x})^{2n}\\
&=\sum_{n=1}^\infty \frac{H_{n}}{n}(\sqrt{x})^n(1+(-1)^n)\\
&=\operatorname{Li}_2(\sqrt{x})+\frac12\ln^2(1-\sqrt{x})+\operatorname{Li}_2(-\sqrt{x})+\frac12\ln^2(1+\sqrt{x})\\
&=\frac12\operatorname{Li}_2(x)+\frac12\ln^2(1-\sqrt{x})+\frac12\ln^2(1+\sqrt{x})
\end{align}
Thus 
$$\boxed{S_4=\frac1x\left(\ln^2(1-\sqrt{x})+\ln^2(1+\sqrt{x})-\frac12\ln^2(1-x)+2\ln2\ln(1-x)\right)}$$

I hope someone will take care of $S_1$ and I hope you find my attempt helpful. 
