Infinite sum including Gaussian Hypergeometric Function

We use the integral representation for the hypergeometric function: \begin{equation} \mathbf{F}\left(a,b;c;z\right)=\frac{1}{\Gamma\left(b\right)\Gamma\left(c-b\right)}\int_{0}^{1}\frac{t^{b-1}(1-t)^{c-b-1}}{(1-zt)^{a}}\mathrm{d}t, \end{equation} where $\mathbf{F}\left(a,b;c;z\right)$ is the regularized hypergeometric function, $\arg\left|1-z\right|<\pi$, $\Re c>\Re b>0$. With $a=1+i,b=i-b,c=1+i-b, z=-x$, the above identity can be used, thus \begin{align} S&=\sum_{i=1}^{+\infty} \frac{ x^{i}}{i-\beta} {_{2} F_{1}}\left(i+1, i-\beta ; i+1-\beta ;-x\right)\\ &=\sum_{i=1}^{+\infty} \frac{ x^{i}}{i-\beta}\frac{\Gamma(i+1-\beta)}{\Gamma(i-\beta)}\int_{0}^{1}\frac{t^{i-\beta-1}}{(1+xt)^{i+1}}\mathrm{d}t \end{align} After simplification of the Gamma factors and change of the order summation/integration, we have \begin{align} S&=\int_0^1\frac{t^{-\beta-1}}{1+xt}\mathrm{d}t\sum_{i=1}^{+\infty}\left( \frac{xt}{1+xt} \right)^i\\ &=x\int_0^1\frac{t^{-\beta}}{1+xt}\mathrm{d}t \end{align} From the above integral representation, with $a=1,b=1-\beta,c=2-\beta,z=-x$, it comes \begin{align} S&=x\,\frac{\Gamma(1-\beta)}{\Gamma(2-\beta)} {_{2}F_{1}}\left(1,1-\beta ; 2-\beta ;-x\right)\\ &= \frac{ x }{1-\beta} {_{2}F_{1}}\left(1,1-\beta ; 2-\beta ;-x\right) \end{align} as expected.

Using the same method for the second identity, we obtain \begin{align} T&=\sum_{k = 1}^{\infty} \frac{(U)_k }{ k! } \frac{\beta x^k}{k - \beta} {_2F_{1}} \left(U + k, k-\beta; k- \beta +1;-x \right) \\ &=\sum_{k = 1}^{\infty} \frac{(U)_k }{ k! } \frac{\beta x^k}{k - \beta}\frac{\Gamma(k-\beta+1)}{\Gamma(k-\beta)}\int_0^1\frac{t^{k-\beta-1}}{(1+xt)^{U+k}}\,dt\\ &=\beta\int_0^1\frac{t^{-\beta-1}}{(1+xt)^U}\,dt\sum_{k = 1}^{\infty} \frac{(U)_k}{k!}\left( \frac{xt}{1+xt} \right)^k \end{align} But \begin{align} (U)_k&=(-1)^k(-U-k+1)_k\\ &=(-1)^k\frac{\Gamma(-U+1)}{\Gamma(-U-k+1)}\\ \frac{(U)_k}{k!}&=(-1)^k\frac{\Gamma(-U+1)}{\Gamma(-U-k+1)\Gamma(k+1)}\\ &=(-1)^k\binom{-U}{k} \end{align} then, with Newton's generalized binomial theorem \begin{align} \sum_{k = 1}^{\infty} \frac{(U)_k}{k!}\left( \frac{xt}{1+xt} \right)^k&=\sum_{k = 0}^{\infty} \binom{-U}{k}\left(-\frac{xt}{1+xt} \right)^k -1\\ &=\left( 1-\frac{xt}{1+xt} \right)^{-U}-1\\ &=(1+xt)^{U}-1 \end{align} Using the generalized binomial theorem one more time, \begin{align} T&=\beta\int_0^1t^{-\beta-1}\left( 1-\frac{1}{(1+xt)^U} \right)\,dt\\ &=\beta\int_0^1t^{-\beta-1}\left[1-\sum_{s=0}^\infty\binom{U+s-1}{s}(-xt)^s\right]\,dt\\ &=-\beta\sum_{s=1}^\infty\binom{U+s-1}{s}(-x)^s\int_0^1t^{s-\beta-1}\,dt\\ &=-\beta\sum_{s=1}^\infty\binom{U+s-1}{s}\frac{(-x)^s}{s-\beta} \end{align} This summation can be written as \begin{align} T&=-\beta\sum_{s=1}^\infty\frac{\Gamma(U+s)}{\Gamma(U)\,s!}\frac{\Gamma(s-\beta)}{\Gamma(s-\beta+1)}(-x)^s\\ &=\sum_{s=1}^\infty\frac{(U)_s(-\beta)_s}{(1-\beta)_s}\frac{(-x)^s}{s!}\\ &=\,_2F_1(U,-\beta;1-\beta;-x)-1 \end{align} which is the proposed identity.