We have by definition of polynomial ring over sequences in $R$ that: :$P \left[{R}\right] = \left\{{\left \langle {r_0, r_1, r_2, \ldots}\right \rangle}\right\}$ where each $r_i \in R$, and all but a finite number of terms is zero. === Proof that Operations are Closed === We need to ensure that the operations as defined are closed. Let $r = \left \langle {r_0, r_1, r_2, \ldots}\right \rangle, s = \left \langle {s_0, s_1, s_2, \ldots}\right \rangle \in P \left[{R}\right]$. As all but a finite number of terms of $r$ and $s$ are zero, there exist $m, n \ge 0$ such that: :$\forall i > m: r_i = 0$ :$\forall j > n: s_j = 0$ Let $l = \max \left\{{m, n}\right\}$. We can express the operations on $P \left[{R}\right]$ as: {{begin-eqn}} {{eqn | n=1       | l=\left \langle {r_0, r_1, \ldots, r_m}\right \rangle + \left \langle {s_0, s_1, \ldots, s_n}\right \rangle       | r=\left \langle {r_0 + s_0, r_1 + s_1, \ldots, r_l + s_l}\right \rangle       | c= }} {{eqn | n=2       | l=-\left \langle {r_0, r_1, \ldots, r_m}\right \rangle       | r=\left \langle {-r_0, -r_1, \ldots, -r_m}\right \rangle       | c= }} {{eqn | n=3       | l=\left \langle {r_0, r_1, \ldots, r_m}\right \rangle \left \langle {s_0, s_1, \ldots, s_n}\right \rangle       | r=\left \langle {t_0, t_1, \ldots, t_{m+n} }\right \rangle       | c=where $\displaystyle t_k = \sum_{i \mathop + j \mathop = k} r_i s_j$ }} {{end-eqn}} We have that: : $\forall i > l: r_i + s_i = 0$ and so  : $r + s \in P \left[{R}\right]$ Equally clearly: : $-r \in P \left[{R}\right]$ Now consider: : $\displaystyle \left({r s}\right)_i = \sum_{j \mathop + k \mathop = i}r_j s_k$ Let $i > m + n$. Then in any $r_j s_k$ such that $j + k = i$, either $j > m$ or k > n. In the first case $r_j = 0$ and in the second $s_j = 0$. In either case $r_j s_k = 0$. So: : $\displaystyle \forall i > m + n: \left({r s}\right)_i = \sum_{j \mathop + k \mathop = i}r_j s_k = 0$ So: : $r s \in P \left[{R}\right]$ {{qed|lemma}} === Proof of Additive Group === The addition operation $r + s$ is clearly commutative and associative, and: :$\left \langle {0, 0, \ldots}\right \rangle = 0_{P \left[{R}\right]}$ Equally clearly: :$\forall r \in P \left[{R}\right]: r + \left({-r}\right) = \left \langle {r_i + \left({-r_i}\right)}\right \rangle = \left \langle {0, 0, \ldots}\right \rangle = 0_{P \left[{R}\right]}$ and so $-r$ is the inverse of $r$ for addition. So $\left({P \left[{R}\right], +}\right)$ is an abelian group, as it needs to be for $P \left[{R}\right]$ to be a ring. {{qed|lemma}} === Proof of Ring Product === We need to establish that the ring product is associative. Let $r = \left \langle {r_0, r_1, \ldots}\right \rangle, s = \left \langle {s_0, s_1, \ldots}\right \rangle, t = \left \langle {t_0, t_1, \ldots}\right \rangle \in P \left[{R}\right]$. Then: {{begin-eqn}} {{eqn | l = \left({\left({r s}\right) t}\right)_n       | r = \sum_{i \mathop + j \mathop = n} \left({r s}\right)_i t_j       | c =  }} {{eqn | r = \sum_{i \mathop + j \mathop = n} \left({\sum_{k \mathop + l \mathop = i} r_k s_l}\right) t_j       | c =  }} {{eqn | r = \sum_{k \mathop + l \mathop + j \mathop = n} r_k s_l t_j       | c =  }} {{end-eqn}} Similarly for $\left({r \left({s t}\right)}\right)_n$. So the ring product is associative, and so forms a semigroup. {{qed|lemma}} === Proof of Distributivity === Finally we need to show that the ring product is distributive over ring addition. {{begin-eqn}} {{eqn | l = \left({\left({r + s}\right) t}\right)_n       | r = \sum_{i \mathop + j \mathop = n} \left({r_i + s_i} \right) t_j       | c =  }} {{eqn | r = \sum_{i \mathop + j \mathop = n} r_i t_j + \sum_{i \mathop + j \mathop = n} s_i t_j       | c =  }} {{eqn | r = \left({r t}\right)_n + \left({s t}\right)_n       | c =  }} {{end-eqn}} Similarly for $\left({t \left({r + s}\right)}\right)_n$. Hence the result. {{qed}}