Simplifying Series of Gamma Function

Question

I am trying to simplify summation of gamma function

$$\sum_{k=1}^\infty\frac{\Gamma(k+b)}{(k-1)!\Gamma(b)} = \sum_{k=1}^\infty\frac{\Gamma(k+b)}{\Gamma(k)\Gamma(b)} , b > 1$$

I need hint/direction here. What is the property that I should use to simplify the expression into a non-summation form?

Thanks!

Answer 1

How about using the definition of the $\Gamma$ function?

$$\begin{eqnarray*}\sum_{k=1}^{+\infty}\frac{\Gamma(b+k)}{\Gamma(b)\Gamma(k)}&=&\frac{1}{\Gamma(b)}\sum_{k\geq 1}\frac{1}{(k-1)!}\int_{0}^{+\infty}t^{k+b-1}e^{-t}\,dt\\&=&\frac{1}{\Gamma(b)}\int_{0}^{+\infty}\sum_{n\geq 0}\frac{t^{n+b}}{n!}e^{-t}\,dt\\&=&\frac{1}{\Gamma(b)}\int_{0}^{+\infty}t^b\,dt\end{eqnarray*} $$ hence the series is never converging. On the other hand: $$\begin{eqnarray*}\sum_{k\geq 1}\frac{\Gamma(k)\Gamma(b)}{\Gamma(b+k)}=\sum_{k\geq 1}B(b,k)&=&\int_{0}^{1}\sum_{k\geq 1}u^{b-1}(1-u)^{k-1}\,du\\&=&\int_{0}^{1}u^{b-2}\,du=\frac{1}{b-1}\end{eqnarray*}$$ provided that $b>1$.

Answer 2

\begin{align} S &= \sum_{k=1}^{\infty} \frac{\Gamma(k+b)}{(k-1)!\Gamma(b)} = \sum_{k=1}^{\infty} \frac{\Gamma(k+b)}{\Gamma(k)\Gamma(b)} \\ &= \sum_{k=0}^{\infty} \frac{b \, \Gamma(k+b+1)}{k! \, \Gamma(b+1)} \\ &= b \, \sum_{k=0}^{\infty} \frac{(b+1)_{k}}{k!} \\ &= b \, (1 - 1)^{-b-1}. \end{align} If $b$ is not negative then this series is infinite. Now, if $b = -a - n$, for $n \geq 1$, then the series is zero.