Let $R$ a ring and we define $$\text{seq}R=\{f=(a_0,a_1,\dots,)\;|\;a_i\in R\}.$$
On this set we define the following operation: let $f=(a_n)_{n\ge0}$ and $g=(b_n)_{n\ge0}$ $$fg=(c_0,c_1,\dots,),$$ where $$c_k=\sum_{i+j=k} a_ib_j.$$
My problem is to prove that this product is associative.
That is let $f=(a_n)_{n\ge0}, g=(b_n)_{n\ge 0}, h=(c_n)_{n\ge 0}$. I must prove that $(fg)h=f(gh)$.
Now $$(fg)h=(d_0,d_1,\dots,)(c_0,c_1,\dots),$$ where $d_k=\sum_{i+j=k} a_ib_j$. Then $$(fg)h=(d_0,d_1,\dots,)(c_0,c_1,\dots,)=(e_0,e_1,\dots),$$ where $$e_s=\sum_{k+t=s}d_kc_t,$$ therefore $$e_s=\sum_{k+t=s}\bigg(\sum_{i+j=k}a_ib_j\bigg)c_t.$$
How can I proceed now to show that $(fg)h=f(gh)$?
Thanks!
Letting $a_n=b_n=c_n=0$ for $n<0$, you can rewrite the formula as $$ e_s=\sum_{k=0}^\infty\biggl(\sum_{i=0}^{\infty}a_ib_{k-i}\biggr)c_{s-k}= \sum_{k=0}^{\infty}\sum_{i=0}^{\infty}a_ib_{k-i}c_{s-k} $$ and now you can switch the sums: $$ e_s=\sum_{i=0}^\infty\sum_{k=0}^{\infty}a_ib_{k-i}c_{s-k}= \sum_{i=0}^\infty a_i\biggl(\sum_{k=0}^{\infty}b_{k-i}c_{s-k}\biggr) $$