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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?

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$\mathbf Q_7$ 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

1 Answer 1

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.

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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

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