At each level, mark the cosets of Hn+1 in Hn. The bottom level is simply R/H. You will need to designate specific cosreps for these cosets, at every level. Let S be the set of infinite sequences of cosets, selecting one cosrep at each level. This will become our inverse limit, but we have some technical details to take care of, like turning S into a ring.
Consider the first n terms of a sequence in S. A correspondence map carries these partial sequences to the elements of R/Hn. Basically, the map adds up S1 through Sn, giving a unique element in R/Hn. The first cosrep S1 specifies a coset of H in R, the second specifies a coset of H2 in H, and so on. Partial seequences can be added or multiplied together by mapping to R/Hn, applying the operation there, and mapping back. Addition looks simple - add the elements Si term by term - and sometimes it is that simple. But remember, you must turn cosreps into cosets, add the cosets, and turn the sum back into its designated cosrep. This may or may not be the sum of the two original cosreps in R.
Good news, the ring properties are inherited from R/Hn. However, we need to prove the sum and product are well defined across the entire sequence.
When you add two cosets of Hn+1 in Hn, this does not change the sum of the cosets at the previous levels. Addition remains consistent for all n.
Multiply two partial sums of length n+1. The last term in each sum is in the ideal Hn, so all the pairwise products that use these terms live in Hn. This does not change the coset of Hn in lesser powers of H. The cosets that have already been established do not change. Thus multiplication is well defined for the entire sequence, and S is a ring.
By construction, a ring homomorphism maps S onto R/Hn - simply restrict attention to the first n terms of the sequences of S.
Apply the aforementioned ring homomorphism from S to R/Hn, then map onto R/Hm, for m < n. Alternatively, go straight from S to R/Hm. The result is the same - the diagram commutes. Therefore S is a limit of the system.
Let T be another limit of the system. For any x in T, map x into each R/Hn, and pull these images back to a sequence in S. You need to prove this map, from T into S, is well defined, is unique, and is a ring homomorphism. I'll leave the details to you. Therefore S is the inverse limit of the system R/Hn.
The inverse limit is unique, up to isomorphism, so a different set of cosreps produces essentially the same ring. There is one completion of R through H, up to isomorphism.
When H is the maximal ideal of a dvr, S is isomorphic to the H-adic completion of R, as a valuation ring. In this case the cosreps are convenient - K times the powers of t, where K is the residue field, and t generates H. Addition and multiplication are also convenient - following the power series rules, with possible carry operations. But the inverse limit can exist for other rings, even rings with zero divisors.
Let J be the intersection of Hn, and let x be a member of J. Select coset after coset after coset, to map x into S. Each x in J leads to the zero sequence in S. Therefore, a ring homomorphism takes R/J into S.
Suppose x, outside of J, maps to 0 in S. Thus x is the 0 coset at each level. This places x inside each Hn, and in their intersection. This is a contradiction, hence R/J becomes a subring of S. If J = 0, e.g. when R is a dvr, R embeds in S.
Let v(x) = m, and v(y) = n, for m < n. Their sum (or difference) lies in Hm. If it lies in Hi, for i > m, then subtract y, and x lies in Hi. This is a contradiction, hence the valuation of the sum is the lesser valuation. If m = n, v(x-y) is at least m. Use these properties to prove symmetry, and the triangular inequality - thus giving a metric space. The completion of this metric space is the completion of R, as an inverse limit. This was described in valuation rings - no point in rehashing it here. We are merely extending these results to other rings and ideals. The only difference is, the topology, and its completion, are restricted to R. We cannnot extend these results to the fraction field of R - in fact R may not even have a fraction field.
In practice, higher ordinals are not used very often. Hensel's three lemmas (presented on the next page), and the theorems that rely on these lemmas, do not extend through a limit ordinal. They only apply to a countable system of quotient rings R/Hn, such as a dvr.
At an intuitive level, we are merely grouping the digits together into blocks of size 2, like reading a number in base 100 instead of base 10. Operations add or multiply 2 digits at a time, but the math is the same.
Technically, you want to build a map from one completion into the other, and prove it is a ring isomorphism. Any system of cosreps will do, so add the first and second cosreps in the first sysstem to get the first cosrep in the second system. In other words, a coset of H in R, plus a coset of H2 in H, becomes a coset of H2 in R. apply this correspondence all the way down the line. Note that it is unique, and reversible. Clearly the math in both systems is the same. Thus we have a ring isomorphism between the two inverse limits.
The completions are also equivalent as topological spaces. When moving from H to H2, you may have to lower the valuation by 1, but this multiplies the distance by 1/c, which is a fixed constant. The map is uniformly bicontinuous, and a homeomorphism.
Bear in mind, you can't base a true valuation on H2; since it does not produce a valuation monoid. You have lost v(xy) = v(x) + v(y). Consider x in H-H2. This has valuation 0 (relative to H2). Yet v(x2) has valuation 1, which is not v(x) + v(x).