given $\mathbb{P}(X=n, Y \leq t)$ find $\mathbb{P}(X=n)$.

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$X, Y$ are random variables

let $a,b > 0$, $ \operatorname{supp}(X,Y) = \mathbb{N} \times \mathbb{R}_{+}$ and

$$\mathbb{P}(X=n, Y \leq t) = \int_0^{t}\frac{(ay)^n}{n!}\exp(-(a+b)y)\,dy$$

my approach consists of finding the joint CDF then taking the limit as $t\to \infty$ to obtain $X$'s CDF and then hoping to end up with a well known CDF, however it failed :

$$\mathbb{P}(X \leq n, Y \leq t) =\sum_{k = 0}^n \mathbb{P}(X = k, Y \leq t)$$

then $$\begin{align}\mathbb{P}(X \leq n) &= \lim_{t\to\infty} \sum_{k = 0}^n \mathbb{P}(X = k, Y \leq t) \\ &=\lim_{t\to\infty}\int_0^{t}\sum_{k = 0}^n \frac{(ay)^k}{k!}\exp(-(a+b)y)\,dy \\ &=\int_0^{\infty}\sum_{k = 0}^n \frac{(ay)^k}{k!}\exp(-(a+b)y)\,dy \\ &=\sum_{k = 0}^n\int_0^{\infty} \frac{(ay)^k}{k!}\exp(-(a+b)y)\,dy \\ &=\sum_{k = 0}^n \frac{a^k}{(a+b)^{k+1}k!}\int_0^{\infty} z^{k}\exp(-z)\,dz\\ &=\sum_{k = 0}^n \frac{a^k}{(a+b)^{k+1}} \end{align}$$

I think I'm doing something wrong because :

$$\lim_{n\to+\infty} \sum_{k = 0}^n \frac{a^k}{(a+b)^{k+1}} =\frac{1}{b} \neq 1$$

what am I doing wrong ? is it because we're dealing with a hybrid pair of random variables ?