Relationship between Poisson and Gamma Distribution There are several solutions to this problem but I am interested in the solution in Casella & Burger Pg. 100. The problem shows that if $X$ follows gamma($\alpha$, $\beta$), a random variable and $Y \sim 
 Poisson(x/\beta)$, then
$P(X\le x) = P(Y\ge \alpha)$.
In the text,
$P(X \le x) = \frac{1} {(\alpha - 2)! \beta^{\alpha - 1}}\int_{0}^{x} t^{\alpha - 2} e^{-t/\beta} \,dt - P(Y = \alpha -1 )$
After second integration by parts (as suggested by the text), I get
$P(X \le x) = \frac{1} {(\alpha - 3)! \beta^{\alpha - 2}}\int_{0}^{x} t^{\alpha - 2} e^{-t/\beta} \,dt  - P(Y = \alpha -2 ) - P(Y = \alpha -1 )$
I surmise that at the $n^{th}$ step I will have
$P(X \le x) = \frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dt  -  P(Y = \alpha -(n-1)) \ldots -P(Y = \alpha -2 ) - P(Y = \alpha -1 )$
which will be
$\frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dt  - [ P(Y = \alpha -(n-1)) \ldots +P(Y = \alpha -2 ) + P(Y = \alpha -1 )]$
$ = \frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dt - [ P(Y = \alpha -(n-1)) \ldots +P(Y = \alpha -2 ) + P(Y = \alpha -1 )]$
$ = \frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dt  -  P(Y \le \alpha- 1) $
I want to argue that as $n \rightarrow \infty $ , $\frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta}= 1 $. In which case,
$ = 1  -  P(Y \le \alpha- 1) = P(Y \ge \alpha) $.
I am not sure however, if $\frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dt$  is an integral of a valid pdf.
Thanks for the help in advance.

You are repeating integration by parts of the term
$$\frac{\int_{0}^{x} t^{\alpha - 1 - n} e^{-t/\beta} \,dt} {(\alpha - 1 - n)! \beta^{\alpha - n}} =  - \left[ \frac{(t/\beta)^{\alpha - 1 - n} e^{-t/\beta}   } {(\alpha - 1 - n)! } \right]_0^x+\frac{\int_{0}^{x} t^{\alpha - 2 - n} e^{-t/\beta} \,dt} {(\alpha - 2 - n)! \beta^{\alpha - 1- n}} $$
The repeated integration by parts will terminate at the point when $n = \alpha - 1$. The last term becomes
$$\frac{1} {(0)! \beta^{1}}\int_{0}^{x} t^{0} e^{-t/\beta} \,dt =  -\left[e^{-t/\beta}\right]_0^x  = 1 - P(Y=0)$$
