Finding the constant of a joint probabiity distribution

Question

Let $A$ and $B$ be random variables such that, for some $\theta>0$,

$P(A=x,B=y)$ = $\theta$$\frac{2x+y}{x!y!}$$(0.33^{x+y})$

for $x=0,1,2,3,…$ and $y=0,1,2,3,….$

How do I find the value of $\theta$?

Current attempt:

$\sum_{x=0}^\infty$ $\sum_{y=0}^\infty$ $\theta$$\frac{2x+y}{x!y!}$$(0.33^{x+y})$ $=1$

Problem: How do I do this summation?

Answer 1

Use the Taylor expansion of the Euler exponential: $$\mathrm e^z =\sum_{k=0}^\infty \frac{z^k}{k!}$$

So...

$$\begin{align}\sum_{x=0}^\infty\sum_{y=0}^\infty\dfrac{(2x+y)0.33^{x+y}}{x!~y!}&=\sum_{x=0}^\infty\left(\dfrac{2x0.33^x}{x!}\sum_{y=0}^\infty\dfrac{0.33^y}{y!}+\dfrac{0.33^x}{x!}\sum_{y=1}^\infty\dfrac{0.33^y}{(y-1)!}\right)\\[1ex]&=\mathrm e^{0.33}\sum_{x=0}^\infty\left(\dfrac{2x0.33^x}{x!}+\dfrac{0.33^{x+1}}{x!}\right)\\[1ex]&=\mathrm e^{0.33}\left(2\cdot0.33\sum_{x=1}^\infty\dfrac{0.33^{x-1}}{(x-1)!}+0.33\sum_{x=0}^\infty\dfrac{0.33^x}{x!}\right)\\[1ex]&=0.99\cdot\mathrm e^{0.66}\end{align}$$