Let $R$ be a ring with the $I$-adic filtration. Recall that a sequence $x_n$ of elements in $R$ is Cauchy if for every $N$, there is a $k$ such that for all $i,j\ge k$ one has $x_i-x_j\in I^N$. A sequence converges to zero if for every $N$, there is a $k$ such that for all $i\ge k$ one has $x_i\in I^N$.
I'd like to understand why the set of all Cauchy sequences forms a ring and why the set of all zero-convergent sequences forms an ideal in that ring.
It is obvious that the two above-mentioned sets are closed under addition. However, it is not clear to me why the first set is closed under multiplication and why the second one is closed under "scalar multiplication".
For example, let $x_n, y_n$ be Cauchy sequences. Then $\forall N$ $\exists k: \forall i,j\ge k, x_i-x_j\in I^N$ and WLOG we may assume that $\forall N$ $\exists k: \forall l,m\ge k, y_l-y_m\in I^N$. Now we for every $N$ we need to find a $K$ such that for all $i,j,l,m\ge K$, $(x_i-x_j)(y_l-y_m)=x_iy_l+x_jy_l-x_iy_m+x_jy_m\in I^N$. But $I^N$ consists of elements of the form $\sum a_1\dots a_n$, where each $a_i$ lies in $I$, and it is unclear why e.g. $x_iy_l\in I^N$. A similar question arises in the case of zero-convergent sequences.
It's not exactly you have to check for multiplication, but $$\forall N\,\exists K\,\forall i, j\ge K,\;x_iy_i-x_jy_j\in I^N.$$ We know there exists $k$, such that $\;\forall i, j\ge k$, $\;x_i-x_j$ and $y_i-y_j\in I^N$. Hence for all such $i,j$, we have $$x_iy_i-x_jy_j=x_i(y_i-y_j)+(x_i-x_j)y_j\in x_i I^N +y_jI^N\subset I^N+I^N=I^N.$$