Evaluating $\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m +a)}{\Gamma(k+m+b)}(k+m)$

Question

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.

Answer 1

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.

Answer 2

Mathematica says (after slight massaging):

$$m \, _2F_1(1,a+m;b+m;1)+(a+m) \Gamma (b+m) \, _2\tilde{F}_1(2,a+m+1;b+m+1;1)$$