Infinite sum including Gaussian Hypergeometric Function How can we prove the following equality?
$$ \sum_{i=1}^{+\infty} \frac{ x^{i}}{i-\beta} {_{2} F_{1}}\left(i+1, i-\beta ; i+1-\beta ;-x\right) = \frac{ x }{1-\beta} {_{2}F_{1}}\left(1,1-\beta ; 2-\beta ;-x\right)  ,$$
where ${_{2}F_{1}}\left(a,b ;c;z \right) $ is the Guassian Hypergeometric Function, and $\beta<1$. (This question was rised when I read the following paper: https://arxiv.org/pdf/1306.6169.pdf , on page 26, the paragraph below the equation (48), where $k_0$ is the right-hand side item and $k_i$ is the $i$-th item for the left-hand side.)
By the way, if it is possible that the following equality also holds? If so, how can we prove it?
$$ \sum_{k = 1}^{\infty}  \frac{  (U)_k }{ k! }  \frac{\beta x^k}{k - \beta}  {_2F_{1}} \left(  U   +   k, k \! - \! \beta;  k \!- \! \beta \!+ \! 1 ; \!-x   \right) = {_2F_{1}}\left(-\beta, U ; 1-\beta ;-x \right)-1,$$
where $U\geq1$, $(U)_k  = U(U+1)\cdots(U+k-1)$ is the rising Pochhammer symbol, and $0<\beta<1$.
 A: 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.
