I want to solve this summation which is given as : $$\sum_{n=1}^{\infty}(\sum_{k=1}^nv_{2k}u_{2n-2k})z^{2n}$$ note that $v_{2k}$ is the first arrival to origin in $2k$ steps and $u_{2n-2k}$ is the arrival to the origin, and $z$ represent the z-transform which is done to find the moment generating function. I substitute $n-k=m$ and got: $$\sum_{m=1-k}^{\infty}u_{2m}z^{2m}\sum_{k=1}^{m+k}v_{2k}z^{2k}$$ I do not know how to move from here to the answer, which is: $$\sum_{m=0}^{\infty}u_{2m}z^{2m}\sum_{k=0}^{\infty}v_{2k}z^{2k}=U(z)V(z)$$ what are the steps that I can't think about This looks like convolution but here the first summation is dependent on the second one!!!
PS: As per the comment by Greg Martin, I assume that the values of outer index $m=n-k$ can vary from $0$ to $\infty$. However for inner integral taking suppose $m=0$ the values can vary from ($k=n-m$), i.e., $1$ to $\infty$, similarly for $m=1$ $k=[0,\infty]$ which is the integer range. Is this how one should approach this. Cn we simply make two of the summations independent of one another like this? Am I correct?