Suppose I have the following product: $$\sum_{k=0}^{a}\alpha_kx^k\sum_{k=0}^{b}\beta_kx^k\sum_{k=0}^{c}\gamma_kx^k$$ Note that the bounds are finite and not equal; $a\neq b \neq c$. I'm looking to simplify this expression so I set about it as follows: Combining the first two summations gives $$\sum_{n=0}^{a+b} \left(\sum_{k=0}^{n}\alpha_k\beta_{n-k}\right)x^k\sum_{k=0}^{c}\gamma_kx^k$$ To help simplify let $\delta_k=\sum_{k=0}^n\alpha_k\beta_{n-k}$. Then $$\sum_{k=0}^{a+b} \delta_k x^k\sum_{k=0}^{c}\gamma_kx^k$$ $$\sum_{n=0}^{a+b+c}\left( \sum_{k=0}^{n}\delta_k\gamma_{n-k} \right)x^n$$ $$\sum_{n=0}^{a+b+c}\sum_{k=0}^{n}\delta_k\gamma_{n-k}x^n$$ Plugging back in for $\delta_k$ gets me $$\sum_{n=0}^{a+b+c}\sum_{k=0}^{n}\sum_{j=0}^k\alpha_j\beta_{k-j}\gamma_{n-k}x^n$$
Would this be the correct representation as a convolution? Are the indices correct and are the bounds properly accounted for?
EDIT: My main concern is this: Let's keep it simple and let $a=2,b=3,c=4, \alpha_k=\beta_k=\gamma_k=\frac1{k!}$. I only really need to show the product of the first two finite sums, so $$\sum_{k=0}^{2}\frac{x^k}{k!}\sum_{k=0}^{3}\frac{x^k}{k!}\sum_{k=0}^{4}\frac{x^k}{k!}$$ By my above argument, the product of the first and second summation would be $$\sum_{n=0}^{5} \left(\sum_{k=0}^{n}\frac1{k!}\frac1{(n-k)!}\right)x^k\sum_{k=0}^{4}\frac1{k!}x^k$$ Looking just at $\sum_{n=0}^{5} \left(\sum_{k=0}^{n}\frac1{k!}\frac1{(n-k)!}\right)x^k$ we get $$\sum_{n=0}^{5} \left(\sum_{k=0}^{n}\frac1{k!}\frac1{(n-k)!}\right)x^k=\frac{1}{0!0!}+\left(\frac{1}{0!1!}+\frac{1}{1!0!}\right)x+\left(\frac{1}{0!2!}+\frac{1}{1!1!}+\frac{1}{2!0!}\right)x^2+\left(\frac{1}{0!3!}+\frac{1}{1!2!}+\frac{1}{2!1!}+\frac{1}{3!0!}\right)x^3+\left(\frac{1}{0!4!}+\frac{1}{1!3!}+\frac{1}{2!2!}+\frac{1}{3!1!}+\frac{1}{4!0!}\right)x^4+\left(\frac{1}{0!5!}+\frac{1}{1!4!}+\frac{1}{2!3!}+\frac{1}{3!2!}+\frac{1}{4!1!}+\frac{1}{5!0!}\right)x^5$$ However, $$\sum_{n=0}^{5} \left(\sum_{k=0}^{n}\frac1{k!}\frac1{(n-k)!}\right)x^k=\left(\frac1{0!}+\frac1{1!}x+\frac1{2!}x^2\right)\left(\frac1{0!}+\frac1{1!}x+\frac1{2!}x^2+\frac1{3!}x^3\right)$$ $$=\frac1{0!0!}+\left(\frac{1}{0!1!}+\frac{1}{1!0!}\right)x+\left(\frac{1}{0!2!}+\frac{1}{1!1!}+\frac{1}{2!0!}\right)x^2+\left(\frac{1}{0!3!}+\frac{1}{1!2!}+\frac{1}{2!1!}\right)x^3+\left(\frac{1}{1!3!}+\frac{1}{2!2!}\right)x^4+\frac{1}{2!3!}x^5$$ Therefore, I don't see how they can be equal. I was hoping for an expression as above but having them be finite sums with different bounds seems to negate the above belief that I had. Can anyone elaborate on this?