Proving that product of 2 pgfs is a pgf - last obstacle

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Let $G_1, G_2$ be probability generating functions of some probability distributions. Prove that $G_1G_2$ is pgf.

What I have:

$G_1=\sum_{n=0}^\infty p_nz^n$ and $G_2=\sum_{k=0}^\infty q_kz^k$ where $\sum_{n=0}^\infty p_n=1$ and $p_n\geq 0$ etc.

Then

$G_1G_2=\sum_{n=0}^\infty p_nz^n\sum_{k=0}^\infty q_kz^k$

It can be easily shown that:

$G_1G_2=\sum_{j=0}^\infty \left[ \sum_{i=0}^j p_iq_{j-i}\right]z^j$

Obviously every $1\geq p_iq_{j-i}\geq0$, but my question is:

How to show that $\sum_{i=0}^j p_iq_{j-i}=1$

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Actually, what you have to show is that $\sum_{j=0}^{\infty}\sum_{i=0}^jp_iq_{j-i}=1$, but there's a way to get around this.

Let $X$ and $Y$ be independent random variables with probability generating functions $G_1$ and $G_2$. Then for each $j\geq 0$, $$ \mathbb{P}(X+Y=j)=\sum_{i=0}^j\mathbb{P}(X=i)\mathbb{P}(Y=j-i)=\sum_{i=0}^jp_iq_{i-j}$$

Therefore $X+Y$ has probability generating function $G_1G_2$, which in particular implies that $G_1G_2$ is a probability generating function.