Polynomial multiplication identity Let $R$ be a nontrivial ring with unity and consider the formal power series ring $R[X]$ over $R$. Let $X$ be the element of $R^{\mathbf{N}}$ for which $X_1=1$ and $X_n=0$ for all $n\ne 1$. Then for every polynomial $p\in R[X]$, there exists a unique $n\in\mathbf{N}$ such that $p_k=0$ for all $n<k$ and
$$p=\sum_{k=0}^np_kX^k.$$
Let $p,q\in R[X]$ be polynomials. I want to prove that
$$\left(\sum_{k=0}^np_kX^k\right)\left(\sum_{j=0}^mq_jX^j\right)=\sum_{k=0}^{n+m}\left(\sum_{j=0}^kp_jq_{k-j}\right)X^k.$$
Attempt:
$$\begin{align*}
\left(\sum_{k=0}^np_kX^k\right)\left(\sum_{j=0}^mq_jX^j\right)&=\sum_{k=0}^n\sum_{j=0}^mp_kX^kq_jX^j\\
&=\sum_{k=0}^n\sum_{j=0}^mp_kq_jX^{j+k}. \end{align*}$$
I am unable to proceed further than this. I probably have to perform an index shift, but I can't see it. Any hints?
Edit:
Actually I think the identity, as it stands, is incorrect. What should the first summation interval be? $[0,n+m]$?
 A: We use the coefficient of operator $[X^n]$ to denote the coefficient of $X^n$ of a polynomial. We see the polynomials, both at the left-hand and at the right-hand side have degree $n+m$ and consider therefore $q$ with $0\leq q\leq n+m$.

We obtain
\begin{align*}
[X^q]\left(\sum_{k=0}^np_kX^k\right)\left(\sum_{j=0}^mq_jX^j\right)
&=\sum_{k=0}^{q}p_k[X^{k-q}]\sum_{j=0}^mq_jX^j\tag{1}\\
&=\sum_{k=0}^{q}p_kq_{k-q}\tag{2}
\end{align*}

Comment:

*

*In (1) we apply the rule $[X^r]X^sP(X) = [X^{r-s}]P(X)$ of the coefficient of operator. Note the upper index of the outer sum is set to $q$. This is admissible, since in case $q\leq n$ only coefficients less or equal to $q$ can provide a non-zero contribution. In case $q>n$ this is also fine, since then we have $p_k=0$ whenever $n<k\leq q$.

*In (2) we select the coefficient of $X^{k-q}$, which is $0$ in case $k-q>m$ and the wanted sum on the other hand.


Since the degree of the polynomial from the left-hand side of (1) is $m+n$, we conclude from (1) and (2):
\begin{align*}
P(X)&=\left(\sum_{k=0}^np_kX^k\right)\left(\sum_{j=0}^mq_jX^j\right)\\
&=\sum_{q=0}^{m+n}\Big([Z^q]P(Z)\Big)X^q\\
&=\sum_{q=0}^{m+n}\left(\sum_{k=0}^{q}p_kq_{k-q}\right)X^q\\
\end{align*}

A: Hint:
Unless you define the coefficients with an out-of-range index to be zero, you cannot express the product with a straight nested sum.
As the figure shows, the groups of coefficients for the terms with constant exponent form two triangular and one parallelogrammic patterns.

You will need a sum with a linearly increasing number of terms, one with a constant number, and one with a decreasing number. You can also arrange sums where the delimiting indexes are made of $\min/\max$ expressions.

