# 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 $$$$H_p=\int_0^1\frac{1-t^p}{1-t}\,dt$$$$ we have \begin{align} H_n-H_{n-1/2}&=\int_0^1\frac{t^{n-1/2}-t^n}{1-t}\,dt\\ &=\int_0^1\frac{t^{n-1/2}(1-\sqrt{t})}{1-t}\,dt\\ &=2\int_0^1\frac{u^{2n}}{1+u}\,du \end{align} The proposed series can be written as \begin{align} S&=\sum_{n=0}^\infty\frac{(1/2)_n}{n!}(H_n-H_{n-1/2})\\ &=2\sum_{n=0}^\infty\frac{(1/2)_n}{n!}\int_0^1\frac{u^{2n}}{1+u}\,du \end{align} by swapping integration and summation \begin{align} S&=2\int_0^1\frac{du}{1+u}\sum_{n=0}^\infty\frac{(1/2)_n}{n!}u^{2n}\\ &=2\int_0^1\frac{du}{(1+u)\sqrt{1-u^2}}\\ &=\int_0^1v^{-1/2}\,dv\\ &=2 \end{align} (the last integral was obtained by changing $$u=(1-v)/(1+v)$$).
• I would have to say $I(1,3/2,1,-1/2)$ is the most complicated way to write the number one I've ever seen! Jan 2 at 19:13