Why does $\sum_{k,l\ge 0} \frac{(k+l)!}{k!l!} \left| x^ky^l\right| =\sum_{n=0}^{\infty} (|x|+|y|)^n$, when $n=k+l$?

Question

Why does $\sum_{k,l\ge 0} \frac{(k+l)!}{k!l!} \left| x^ky^l\right| =\sum_{n=0}^{\infty} (|x|+|y|)^n$, when $n=k+l$?

This was given as fact to see that $\sum_{k,l\ge 0} \frac{(k+l)!}{k!l!} \left| x^ky^l\right| $ converges when $|x|+|y|<1$, but I don't see how it can be rearranged in this way?

Answer 1

$$\sum_{k,l\ge 0} \frac{(k+l)!}{k!l!} \left| x^ky^l\right| $$ Let $ n = k + l$ $$=\sum_{n\ge0}\sum_{l\ge 0}^{n} \frac{n!}{(n-l)! l!}|x^{n-l}y^l|$$ $$=\sum_{n\ge0}(|x| + |y|)^n$$

Answer 2

It is just the binomial theorem, $(a+b)^n =\sum_{k=0}^n \binom{n}{k} a^k b^{n-k} $ applied to the right side and rearranged.