Evaluating $\sum_{n=0}^\infty\frac{(1/2)_n}{n!}(H_n-H_{n-1/2})$

I am seeking a closed-form (a form in terms of known special functions) to the sum $$ \sum_{n=0}^\infty\frac{(1/2)_n}{n!}(H_n-H_{n-1/2}). $$

Context:

I am searching for closed-forms to special cases of the double integral $$ I(\alpha_1,\alpha_2,\zeta,\nu)=\nu^2\int_0^1\int_0^1 {_2F_1}\left({1+\nu,1+\nu \atop \alpha_1};\frac{(1-x)(1-y)}{(1-(1-\zeta)x)(1-(1-\zeta)y)}\right) {_2F_1}\left({1-\nu,1-\nu \atop \alpha_2};(1-x)(1-y)\right)\left((1-(1-\zeta)x)(1-(1-\zeta)y\right)^{-\nu-1}\,\mathrm dx\mathrm dy, $$ which converges for $\alpha_1+\alpha_2>2$, $-\alpha_2<2\nu<\alpha_1$. Also, we require $\zeta\geq 0$, $\alpha_1,\alpha_2>0$, and $\nu\in\Bbb R$. This double integral represents the second moment of a continuous random variable I've been studying.

Substituting $\alpha_1=1$, $\alpha_2=3/2$, $\zeta=1$, and $\nu=-1/2$, I was able to reduce this double integral to $$ I(1,3/2,1,-1/2)=\frac{1}{2}\sum_{n=0}^\infty\frac{(1/2)_n}{n!}(H_n-H_{n-1/2}), $$ where $H_z$ is the Harmonic number. I can provide details to this calculation if desired. But how to proceed? The difference of the harmonic numbers can be reduce to the form $H_n-H_{n-1/2}=a_n+\log 4$, which may be helpful.

From the integral representation \begin{equation} H_p=\int_0^1\frac{1-t^p}{1-t}\,dt \end{equation} we have \begin{align} H_n-H_{n-1/2}&=\int_0^1\frac{t^{n-1/2}-t^n}{1-t}\,dt\\ &=\int_0^1\frac{t^{n-1/2}(1-\sqrt{t})}{1-t}\,dt\\ &=2\int_0^1\frac{u^{2n}}{1+u}\,du \end{align} The proposed series can be written as \begin{align} S&=\sum_{n=0}^\infty\frac{(1/2)_n}{n!}(H_n-H_{n-1/2})\\ &=2\sum_{n=0}^\infty\frac{(1/2)_n}{n!}\int_0^1\frac{u^{2n}}{1+u}\,du \end{align} by swapping integration and summation \begin{align} S&=2\int_0^1\frac{du}{1+u}\sum_{n=0}^\infty\frac{(1/2)_n}{n!}u^{2n}\\ &=2\int_0^1\frac{du}{(1+u)\sqrt{1-u^2}}\\ &=\int_0^1v^{-1/2}\,dv\\ &=2 \end{align} (the last integral was obtained by changing $u=(1-v)/(1+v)$).