Let both $a$,$b$ and $\theta$ be real numbers not equal to a negative integer. Let $m$ be a positive integer. I have shown that the following equality holds:
\begin{eqnarray}
&&S^{a}\_{b,\theta} (m) := \sum\limits\_{i=0}^{m-1} \binom{i+a}{b} \frac{1}{i+\theta} = \\
&&\sum\limits\_{l=1}^\infty \binom{a-\theta}{b-l} \frac{1}{l} \left(\binom{\theta-1+m}{l} - \binom{\theta-1}{l}\right) + \\&&\binom{a-\theta}{b} \left(\psi(\theta+m) - \psi(\theta)\right) =\\
&&\binom{a-\theta}{b-1} \left[
(\theta+m-1) F\_{4,3}\left[\begin{array}{rrrr} 1 & 1 & 1-b & 2-m-\theta \\ 2 & 2 & 2+a-b-\theta &\end{array};1\right]
 - 
(\theta-1) F\_{4,3}\left[\begin{array}{rrrr} 1 & 1 & 1-b & 2-\theta \\ 2 & 2 & 2+a-b-\theta &\end{array};1\right]
\right]
+\\
&&\binom{a-\theta}{b} \left(\psi(\theta+m) - \psi(\theta)\right) 
\end{eqnarray}
Here $\psi$ is the polygamma function of order zero. Now there is a question. What happens if the binomial factor in the sum is replaced by a product of two different binomial factors each one with different values of $a$ and $b$.