Q:

Solving $\sum_{n=y}^{\infty} {n\choose y}\left(\frac{(1-p)\beta}{\beta + 1}\right)^{n}\frac{\Gamma(n + \alpha)}{n!}$

Distribution of $Y$ given $Y\mid N \sim \operatorname{Binomial}(N, p)$, $N\mid \Lambda \sim \operatorname{Poisson}(\Lambda)$ and $\Lambda \sim \operatorname{Gamma}(\alpha, \beta)$.

$$f_N(n) = \int_0^{\infty} f_{N, \Lambda}(n, \lambda)\, d\lambda = \int_0^{\infty} f_{N\mid \Lambda}(n\mid  \lambda)f_{\Lambda}(\lambda) \,d\lambda\\= \dfrac{1}{\Gamma(\alpha) \beta^\alpha n!}\int_0^{\infty} \lambda^{n+\alpha - 1} \exp\left( \dfrac{-\lambda}{\beta/(\beta + 1)}\right) d\lambda$$
Using the substitution $t = \dfrac{\lambda}{\beta/(\beta + 1)}$,
$$f_N(n) = \dfrac{\Gamma(n + \alpha) \beta^n}{\Gamma(\alpha)(\beta + 1)^{n + \alpha}n!}$$ for $n = 0, 1, 2,...$
Now,
$$f_Y(y) = \sum_{n=y}^{\infty} f_{Y,N}(y, n) = \sum_{n=y}^{\infty} f(y\mid n)f_N(n) = \sum_{n=y}^{\infty} {n\choose y}p^y(1-p)^{n-y}\dfrac{\Gamma(n + \alpha) \beta^n}{\Gamma(\alpha)(\beta + 1)^{n + \alpha}n!} \\= \left(\dfrac{p}{1-p}\right)^y\dfrac1{\Gamma(\alpha)(\beta + 1)^{\alpha}}\sum_{n=y}^{\infty} {n\choose y}\left(\dfrac{(1-p)\beta}{\beta + 1}\right)^{n}\dfrac{\Gamma(n + \alpha)}{n!}$$

How should I continue from here on?   

A:

Using the kernels is often useful when working with hierarchical models.  We have
$$f_{\Lambda}(\lambda) \propto \lambda^{\alpha-1} e^{-\lambda/\beta} \\
\Pr[N = n \mid \Lambda = \lambda] \propto e^{-\lambda} \frac{\lambda^n}{n!} \\
$$
hence the unconditional PMF of $N$ is $$\Pr[N = n] \propto \frac{1}{n!} \int_{\lambda=0}^\infty \lambda^{n+\alpha-1} e^{-(1+1/\beta)\lambda} \, d\lambda = \frac{\Gamma(n+\alpha)}{n!(1+1/\beta)^{n+\alpha}}$$
which is the kernel of a negative binomial distribution with parametrization $$\Pr[N = n] = \binom{n+\alpha-1}{\alpha-1} \left(\frac{1}{1+\beta}\right)^\alpha \left(1 - \frac{1}{1+\beta}\right)^n, \quad n \in \{0, 1, 2, \ldots\}$$ or $$N \sim \operatorname{NegativeBinomial}(r = \alpha, \pi = 1/(1+\beta)).$$  So far, this agrees with your calculation.  Then the next step of the hierarchy is
$$\Pr[Y = y \mid N = n] = \binom{n}{y} p^y (1-p)^{n-y},$$ hence 
$$\begin{align*}
\Pr[Y = y] &= \left(\frac{p}{1-p}\right)^y \left(\frac{1}{1+\beta}\right)^\alpha \sum_{n=y}^\infty \binom{n}{y} (1-p)^n \binom{n+\alpha-1}{\alpha-1} (1+1/\beta)^{-n} \\
&= \left(\frac{p}{1-p}\right)^y \left(\frac{1}{1+\beta}\right)^\alpha \binom{y+\alpha-1}{\alpha-1} \sum_{n=y}^\infty \binom{n+\alpha-1}{n-y} \left(\frac{1-p}{1+1/\beta}\right)^n \\
&= \left(\frac{p}{1-p}\right)^y \left(\frac{1}{1+\beta}\right)^\alpha \binom{y+\alpha-1}{\alpha-1} \left(\frac{1-p}{1+1/\beta}\right)^y \left(\frac{1/\beta+p}{1+1/\beta}\right)^{-a-y} \\
&= \binom{y+\alpha-1}{\alpha-1} \left(\frac{1}{1+\beta p}\right)^\alpha \left(\frac{\beta p}{1+\beta p} \right)^y, \quad y \in \{0, 1, 2, \ldots\}.
\end{align*}$$
This is a negative binomial distribution with parameters $$r = \alpha, \quad \pi = 1/(1+\beta p).$$
At one point in our calculation, we used the formula for the binomial series
$$(1-z)^{-(m+1)} = \sum_{k=0}^\infty \binom{k+m}{k} z^k.$$  This occurs in the third step of the derivation of $\Pr[Y = y]$ above, with the choice $k = n-y$, $m = y+ \alpha-1$, and $z = \frac{1-p}{1+1/\beta}$, and is where you had stopped in your calculation.

