How can I multiply these sums?

Question

How do I multiply this? Is it even possible given that one sum is infinite?

$$ x^{n}\left[\,\sum_{r = 0}^{n}\left(-1\right)^{r}x^{mr}\,\right] \left[\,\sum_{k = 0}^{\infty}{n + k - 1 \choose k} x^{k}\,\right] $$

Answer 1

You can collect the like powers of $x^k$ like so: $$ \begin{align} x^n\left[\sum_{r=0}^n(-1)^rx^{mr}\right]\left[\sum_{k=0}^\infty \binom{n+k-1}{k} x^k\right] &=\sum_{k=n}^\infty\left[\sum_{r=0}^n(-1)^r\binom{k-mr-1}{k-mr-n}\right]x^k\\ &=\sum_{k=n}^\infty\left[\sum_{r=0}^{\min\left(\left\lfloor\frac{k-n}{m}\right\rfloor,n\right)}(-1)^r\binom{k-mr-1}{n-1}\right]x^k\\ \end{align} $$ noting that $\displaystyle\binom{n}{k}=0$ when $k\lt0$.

Answer 2

The first factor with the finite alternating geometric sum over $r$ is $$\frac{1-(-x^m)^{n+1}}{x^m+1}$$. The second infinite sum over $k$ is $\frac{1}{(1-x)^n}$.