# Proving a sequence is a cauchy sequence in for the 7 -adic metric

1. May 8, 2013

1. The problem statement, all variables and given/known data
Show that the sequence (xn)n$\in$N $\in$Z given by xn = Ʃ from k=0 to n (7n) for all n $\in$ N is a cauchy sequence for the 7 adic metric.

2. Relevant equations
In a metric space (X,dx) a sequence (xn)n$\in$N in X is a cauchy sequence if for all ε> 0 there exists some M$\in$N such that dx(xn,xm)<ε for all m,n ≥ M.

the 7-adic metric is defined as follows:
p(m,n)= 1/(the largest power of 7 dividing m-n) if m$\neq$n or 0 if m=n

3. The attempt at a solution

I am struggling with proving sequences are cauchy because I am not sure how to go about finding the 'M'? I am not even sure how to start the question apart from assuming m<n. Just a hint at how to start it or how to approach the question would be appreciated.

Assuming m<n am I able to write d7(xn,xm)=Ʃ from k=m+1 to n (7n)?

Thank you.

2. May 8, 2013

### Fredrik

Staff Emeritus
So to calculate $p(x_n,x_m)$, you need to find the largest power of 7 that divides $x_n-x_m$?

That sum at the end is $x_n-x_m$, right? Can you take $7^{m+1}$ outside of the sum, and then do something fun with the sum you have left.
$$x_n-x_m=\sum_{k=m+1}^n 7^k=7^{m+1}(1+7+\cdots+7^{n-m-1}).$$

Last edited: May 8, 2013