I am trying to understand Fraleigh's proof of the fact that the set $R[x]$ of all polynomials in an indeterminate $x$ with coefficients in a ring $R$ obeys the associative law for multiplication. Here are the first few steps of the proof:
Applying ring axioms to $a_i, b_j, c_k \in R$, we obtain:
$\begin{align} \left[\left(\sum_{i = 0}^{\infty}a_ix^i\right)\left(\sum_{j = 0}^{\infty}b_jx^j\right)\right](\sum_{k = 0}^{\infty}c_kx^k) &= \left[\sum_{n = 0}^{\infty} \left(\sum_{i = 0}^{n}a_i b_{n-i}\right)x^n\right](\sum_{k = 0}^{\infty}c_kx^k)\\ &= \sum_{s=0}^{\infty}\left[ \sum_{n = 0}^{s} \left(\sum_{i = 0}^{n}a_i b_{n-i}\right)c_{s-n} \right]x^s \\ &= \sum_{s=0}^{\infty}\left[ \sum_{i + j +k = s}a_ib_jc_k \right]x^s \end{align}$
I have not added the entire proof because I am not even getting the first three steps and I wish to understand the rest of the proof on my own. I think, primarily, I am having trouble understanding how all the summations are working out in the proof. For instance, where does the $n$ index in the first equality of the proof come from? I also dont understand the ensuing equalities. Can someone please explain this proof?
I am aware of this and this, which are similar questions, but they don't really address my question.
What Fraleigh has done is to write the coefficient of the $n$-th term of the product, in terms of the coefficients of the original power series. Then he does it one more time. This is an example of the Cauchy product of two series.
The formula is $(\sum a_ix^i)(\sum b_jx^j)=\sum c_nx^n$, where $c_n=\sum_{i+j=n}a_ib_j$.
Thus we see that there's a typo. It should be $\sum a_ib_{n-\color{red}{i}}$ in the expressions after the first and second equalities. The summation indices need to be fixed too.