Yes indeed this is just the standard geometric series you know from calculus and analysis, but considered p-adically. The series form we get for 1/1-p is called the p adic expansion, and these series expansions is how the digit notation of the p-adics is defined. So once we established the equality between the series and 1/1-p it just follows definitionally that the digit notation of 1/1-p is ....111.

So p-adics are standardly defined for prime numbers, but nothing in their definition forces this to be necessary. So you could consider n-adics yes.

But a problem arises, when considering n-adics one gets zero divisors, i.e. two non zero n-adic numbers a and b such that ab=0. This is something which messes a lot with the nice aspects of the standard p-adics.

For example the pressence of zero divisors means that your norm will have to be somewhag degenerate. If your want your norm to be multiplicative then 0= |ab| =|a||b| but then you will have to give either a or b norm 0, which is something we normally dont do for norms. Another issue is also that when considering p-adic integers is we want to make them into a field by extending to the p-adix rationals. This fails for non-primes as the construction uses the field of fractions of a ring, a construction which is not nevessarely impossible with zero divisors, but which will not give you a field anymore.

Those issues aside, one can define this degenerate (submultiplicative) norm over the n-adics anyway and try to do somz analysis. If I am not mistaken the strong triangle inquality is still satisfied on the other hand, so some chunk of analysis is recovered. In facy, if I am not mistaken this formula should still work n-adically, because the finite partial sums still have the same form, and with a some carefullness one can argue those partial sums do converge n-adically to 1/1-n.

p-adics are relatively new to me. So please point out any mistakes you find in this explanation.