Some Sum's

Due to Euler we have $_{2}F_1(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-zt)^n}dt$ and hence we have $_{2}F_1\left(-n,\frac{1}{2};1;1\right)=\frac{\Gamma(1)}{\Gamma^2\left(\frac{1}{2}\right)}\int_0^1t^{-1/2}(1-t)^{-1/2}(1-t)^ndt$ and required sum is $\sum_{n\geq 1} \frac{_{2}F_1\left(-n,\frac{1}{2};1;1\right)}{n}=-\frac{1}{\pi}\int_{0}^1t^{-1/2}(1-t)^{-1/2}\log t dt$ since $\displaystyle \sum_{n\geq 1} n^{-1}(1-t)^n=-\log t$ for all $t\in\mathbb (0,1]$ and last integral is easy to observe that $\lim_{x\to 1/2}\frac{\partial }{\partial x} \int_0^1t^{x-1}(1-t)^{-1/2}dt=\lim_{x\to 1/2}\frac{\partial}{\partial x}B\left(x,\frac{1}{2}\right)=\lim_{x\to 1/2}B\left(x,\frac{1}{2}\right)\left(\psi_0(x)+\psi_0\left(x+\frac{1}{2}\right)\right)=\pi\left(\psi_0(1)+\psi_0\left(\frac{1}{2}\right)\right)=\pi\left(\gamma-\gamma-2\ln 2\right)=2\ln2$ and thus the final answer we have is $2\ln 2$ giving us $a+b=4$ .

Alternatively

Due to Guass we have $_{2}F_1(a,b:c,1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$ so that $\sum_{n=1}^{\infty}\frac{_{2}F_1\left(-n,\frac{1}{2};1;1\right)}{n}=\sum_{n=1}^{\infty}\frac{\Gamma(1)\Gamma\left(\frac{1}{2}+n\right)}{\Gamma(1+n)\Gamma\left(\frac{1}{2}\right)}$ since $\displaystyle\Gamma\left(n+\frac{1}{2}\right)=\frac{(2n)!}{4^nn!}\sqrt{\pi}$ so the final sum becomes $\sum_{n=1}^{\infty}\frac{1}{n4^n}{2n\choose n}=\int_0^1\left(\frac{1}{x\sqrt{1-x}}-\frac{1}{x}\right)dx=2\ln 2$ _

It is required to specify that either $a$ or $b$ is prime for unique answer since $2\ln(2)=\ln(4)$ gives different answer. For $n\geq 1$ the result can also be written as $\displaystyle \frac{1}{n}\ln(2^{2n})$ so it doesn't make answer unique