Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to do the following question

Find a p-adically Cauchy sequence which converges p-adically to $-1/6$ in $\mathbb{Z}_7$.

In general in $\mathbb{Q}_p$ what is the stronger condition, to be p-adically convergent or p-adically Cauchy?

share|cite|improve this question
$\mathbf Q_7$ is complete (hence so is the closed subset $\mathbf Z_7$), so being convergent and being Cauchy are the same thing. Also, do you want a sequence of integers (to be precise, elements of $\mathbf Z$) which converges to this limit? Otherwise, I don't understand the question. – Dylan Moreland Apr 26 '12 at 0:25
Yes I think that is what I want. The answer I have is $1 + 7 + 7^2 + 7^3 + 7^4 \dots$. – Alex Kite Apr 26 '12 at 0:35
Did you leave out the $7^1$ term on purpose? Assuming it's an accident, your answer looks good, since I multiply it by 6 and get -1. (The sequence you're after is just the successive truncations of your series). – user29743 Apr 26 '12 at 0:37
I did not mean to leave it out! – Alex Kite Apr 26 '12 at 0:38
up vote 2 down vote accepted

I guess the thing to notice is that in the $p$-adics we have a geometric series \[ \frac{1}{1 - p} = 1 + p + p^2 + \cdots; \] you can prove this as before, noting that $|p| < 1$ in our new absolute value. With $p = 7$ the left side is $-1/6$, so the partial sums of the right side form a sequence in $\mathbf Z$ which converges to that number.

share|cite|improve this answer
Alternatively, if you want to find the $p$-adic expansion of an element of the localization $\mathbf Z_{(p)}$, then you really can just look at the image of that element in $\mathbf Z/p\mathbf Z$, $\mathbf Z/p^2\mathbf Z$, etc. This might be better if the number you want to approximate isn't part of a well-known formula. Let me know if you want me to say more about this. – Dylan Moreland Apr 26 '12 at 0:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.