Compute Limit of p-adic Cauchy Sequence

Question

This has really been irking at me and I really should be able to do this but for some reason I can't so I'll ask on here. It is easy to compute the rational "equivalent" of a Cauchy sequence of the form \[ p^{0k} + p^{1k} + p^{2k} + p^{3k} + \cdots \] in $\mathbb{Q}_p$ by scaling each term by $p - 1$ and adding on the sequences produced by multiplying by $p, p^2, \dots, p^{k - 1}$ to get some nice form for $-1$. However, no matter how hard I try, I can't seem to find an analogue for this if we release the restriction of $p$ prime (e.g. $1 + 6^2 + 6^4 + \cdots$) and change the field to be $\mathbb{Q}_{p'}$ where $p'$ is not necessarily related to $p$. So basically, how does one compute a the limit of \[ 1 + 6^2 + 6^4 + \cdots \] in, say, the field $\mathbb{Q}_2$?

Answer 1

$$1+\sum_{n=1} p^{nk}=1+\sum_{n=1} (p^k)^{n}=\sum_{n=0}^m (p^k)^{n}=\frac{p^{k(m+1)}-1}{p^k-1}$$ in your example you have $p=6$ and $k=2$ so we have $$\frac{36^{(m+1)}-1}{35}$$

So what number can be subtract from this such that it gives a power of two number? $-\frac{1}{35}$ works as we get

$$\frac{36^{(m+1)}-1}{35}+\frac{1}{35}=\frac{36^{(m+1)}}{35}=\frac{2^{2(m+1)}3^{2(m+1)}}{35}$$ at which we get for $\mathbb{Q}_2$ $$|\sum_{n=0}^m (6^2)^{n}+\frac{1}{35}|=2^{-2(m+1)}$$ which we can do normal cauchy etc and get it converges to $-\frac{1}{35}$