Answer
$p_{X+Y}(w) = \dfrac{e^{-(\lambda+\mu)}(\lambda+\mu)^w}{w!},\ w=0,1,2,3,\ldots$
See explanation.
Work Step by Step
Since $X$ and $Y$ are Poisson with respective parameters $\lambda$ and $\mu$, then
$p_X(x) = \dfrac{e^{-\lambda}\lambda^x}{x!},\ x=0,1,2,3,\ldots$ and
$p_Y(y) = \dfrac{e^{-\mu}\mu^y}{y!},\ y=0,1,2,3,\ldots$.
If we let $W=X+Y$, then from the first part of Theorem 3.8.3,
$\begin{align*}
p_W(w) &= \sum_{x=0}^\infty p_X(x)p_Y(w-x),\ x=0,1,2,3,\ldots;\ w-x=0,1,2,3,\ldots \\
&= \sum_{x=0}^w p_X(x)p_Y(w-x),\ w=0,1,2,3,\ldots \\
& \qquad\qquad\qquad\qquad\qquad \text{since}\ p_Y(w-x)=0\ \text{for}\ w-x \lt 0 \\
&= \sum_{x=0}^w \dfrac{e^{-\lambda}\lambda^x}{x!}\cdot\dfrac{e^{-\mu}\mu^{w-x}}{(w-x)!} \\
&= e^{-\lambda}e^{-\mu}\sum_{x=0}^w \dfrac{\lambda^x}{x!}\cdot\dfrac{\mu^{w-x}}{(w-x)!} \\
&= \dfrac{e^{-\lambda}e^{-\mu}}{w!}\sum_{x=0}^w \frac{w!}{x!(w-x)!}\lambda^x\cdot\mu^{w-x} \\
&= \dfrac{e^{-\lambda-\mu}}{w!}\underbrace{\sum_{x=0}^w {w\choose x} \lambda^x\cdot\mu^{w-x}}_{(\lambda+\mu)^w} \\
p_W(w) &= \dfrac{e^{-(\lambda+\mu)}(\lambda+\mu)^w}{w!},\ w=0,1,2,3,\ldots.
\end{align*}$
It can be seen that the pdf of $W=X+Y$ is that of a Poisson distribution with rate parameter $\lambda +\mu$.
Thus, $X+Y$ is Poisson with rate parameter $\lambda+\mu$.
