The expected number of events for a Poisson distribution under a Gamma prior with parameters $\alpha$ and $\beta$ (with mean $\alpha/\beta$) is:
$$ \newcommand{\paren}[1]{\left(#1\right)} \begin{aligned} \sum_{k=1}^\infty k & \cdot P(k \mid \alpha, \beta) \\ = \sum_{k=1}^\infty k & \int_0^\infty P(k \mid \lambda) p(\lambda \mid \alpha, \beta) \, d\lambda \\ = \sum_{k=1}^\infty k & \int_0^\infty \paren{\frac{\lambda^k e^{-\lambda}}{k!}} \paren{\frac{\beta^\alpha}{\Gamma(\alpha)} \lambda^{\alpha-1} e^{-\beta \lambda}} \> d\lambda \\ = \sum_{k=1}^\infty k & \> \frac{\beta^\alpha}{k! \> \Gamma(\alpha)} \int_0^\infty \lambda^{k+\alpha-1} e^{-(\beta+1)\lambda} \> d\lambda \\ = \sum_{k=1}^\infty k & \> \frac{\beta^\alpha}{(\beta+1)^{\alpha+k}} \frac{\Gamma(k+\alpha)}{\Gamma(\alpha) \Gamma(k+1)} \\ = \sum_{k=1}^\infty & \frac{\beta^\alpha}{(\beta+1)^{\alpha+k}} \frac{\Gamma(k+\alpha)}{\Gamma(\alpha) \Gamma(k)} \\ = \sum_{k=1}^\infty & \frac{\beta^\alpha}{(\beta+1)^{\alpha+k}} \frac{1}{B(\alpha, k)} \\ = \frac{\beta^\alpha}{(\beta+1)^\alpha} & \sum_{k=1}^\infty \frac{(\beta+1)^{-k}}{B(\alpha, k)} \\ = \paren{\frac{\beta}{\beta+1}}^\alpha & \sum_{k=1}^\infty \frac{(\beta+1)^{-k}}{B(\alpha, k)} \end{aligned} $$ where $\Gamma$ is the Gamma function and $B$ is the Beta function.
Is there any way to further simplify this expression?
Intuitively, I would expect it to be at or near the expected number of events for a Poission distribution with mean equal to the expected $\lambda$ from the prior,
$$ \begin{aligned} & \mathbb{E}_k \text{Poisson}\left(\lambda = \mathbb{E}_\lambda(\text{Gamma}(\alpha, \beta))\right) \\ = {} & \> \mathbb{E}_k \text{Poisson}\left(\lambda = \alpha/\beta\right) \\ = {} & \> \alpha/\beta \end{aligned} $$
Edit
I found a description of the Gamma-Poisson
distribution, which confirms my intuition that the expected value is in fact $\alpha/\beta$.