<s>[INST] Evaluating $\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m +a)}{\Gamma(k+m+b)}(k+m)$ I'm trying to solve the following sum for a project. Many thanks in advance for answering the question!
$$\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m+a)}{\Gamma(k+m+b)}(k+m)$$
where $m$ is an integer but $a$ and $b$ can be any real numbers.
 [/INST] Using this result for hypergeometric functions of unity argument as well as the property of the gamma function $x\Gamma(x)=\Gamma(x+1)$ we find:
$$
\begin{align}
S
&=\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m
+a)}{\Gamma(k+m+b)}(k+m)\\
&=\sum_{k=0}^{\infty}\frac{(m
+a)_k}{(m+b)_k}(k+m)\\
&=\sum_{k=0}^{\infty}\frac{(1)_k(m
+a)_k}{(m+b)_k\,k!}(k+m)\\
&=m\sum_{k=0}^{\infty}\frac{(1)_k(m
+a)_k}{(m+b)_k\,k!}
+\sum_{k=1}^{\infty}\frac{(1)_k(m
+a)_k}{(m+b)_k\,k!}k\\
&=mF(1,m+a;m+b;1)
+\sum_{k=0}^{\infty}\frac{(1)_{k+1}(m
+a)_{k+1}}{(m+b)_{k+1}\,(k+1)!}(k+1)\\
&=m\frac{\Gamma(m+b)\Gamma(b-a-1)}{\Gamma(m+b-1)\Gamma(b-a)}
+\frac{m
+a}{m+b}\sum_{k=0}^{\infty}\frac{(2)_k(m
+a+1)_k}{(m+b+1)_k\,k!}\\
&=m\frac{m+b-1}{b-a-1}
+\frac{m+a}{m+b}F(2,m+a+1;m+b+1;1)\\
&=m\frac{m+b-1}{b-a-1}
+\frac{m+a}{m+b}\frac{\Gamma(m+b+1)\Gamma(b-a-2)}{\Gamma(m+b-1)\Gamma(b-a)}\\
&=m\frac{m+b-1}{b-a-1}
+\frac{(m+a)(m+b-1)}{(b-a-1)(b-a-2)}\\
&=m\frac{m+b-1}{b-a-1}
+\frac{(m+a)(m+b-1)}{(b-a-1)(b-a-2)}\\
&=\frac{m+b-1}{b-a-1}\left(m
+\frac{m+a}{b-a-2}\right),
\end{align}
$$
which holds so long as $\Re(b-a-2)>0$. This result was numerically evaluated in Mathematica and compared with numerical values of the original sum $S$, which showed agreement.
</s>