Sequence in Q with p-diatic metric. Show it converges to a rational

[arturo_026]

This is the problem I'm trying to slove:

Consider the sequence s_n = Sumation (from k=0 to n) p^k (i.e. s_n=p^0+p^1+p^2...+p^n) in Q(rationals) with the p-adic metric (p is prime).
Show that s_n converges to a rational number.[/B]


Now, I do get some intuition on showing that the number it converges to is rational by applying the fundamental theorem of arithmetic and claiming that the number it converges to can be expressed as a product of primes so that way we can factor out p to some power and satisfy the convergence definition with the p-diatic metric.
My big problem is: How can I show that s_n converges in the first place? or should i start my proof by assuming that it converges to some s, and then showing that s is rational?

 

[arturo_026]

something I just thought of is that maybe I can apply a version of the ratio test with the p-adic absolute value in place of the normal absolute value?
So that way the p-adic abs. value of (p^k+1)/(p^k) = p-adic abs. of p which equals to 1/p [by definition of p-adic abs. value]. So this being less than 1, can I conclude that the sequence converges?

 

[Dick]

something I just thought of is that maybe I can apply a version of the ratio test with the p-adic absolute value in place of the normal absolute value?
So that way the p-adic abs. value of (p^k+1)/(p^k) = p-adic abs. of p which equals to 1/p [by definition of p-adic abs. value]. So this being less than 1, can I conclude that the sequence converges?

Do you know a version of the ratio test that applies to p-adic numbers? If not, then I wouldn't just make one up. There's a more straightforward approach. i) show s_n is Cauchy. If so then it converges. Then write a simple formula for s_n and take it's limit.

 

[arturo_026]

Thank you for your response Dick.
So to prove it is Cauchy I've introduced another sequence that has a summation upper limit of m, so call it s_m. And without loss of generality let n>m. now Apropriately choosing e(epsilon) > p^(-m-1) the for m>N, I can have the p-adic absolute value of (s_n - s_m) < epsilon.


I came up with e>p^(-m-1) since p-adic abs. value of (s_n - s_m) = p-adic ((p^0+p^1+...p^n)-(p^0+p^1...+p^m)) = p-adic (p^(m+1) (1+p+...p^(n-m-1))) = p^(-m-1)


Now, to the simple formula, I think you might be referring to is (1-p^(n+1))/(1-p) so I separate this and have 1/(1-p) - p^(n+1)/(1-p) but if I take the limit of the second term, wouldn't it diverge?

 

[Dick]

Thank you for your response Dick.
So to prove it is Cauchy I've introduced another sequence that has a summation upper limit of m, so call it s_m. And without loss of generality let n>m. now Apropriately choosing e(epsilon) > p^(-m-1) the for m>N, I can have the p-adic absolute value of (s_n - s_m) < epsilon.


I came up with e>p^(-m-1) since p-adic abs. value of (s_n - s_m) = p-adic ((p^0+p^1+...p^n)-(p^0+p^1...+p^m)) = p-adic (p^(m+1) (1+p+...p^(n-m-1))) = p^(-m-1)


Now, to the simple formula, I think you might be referring to is (1-p^(n+1))/(1-p) so I separate this and have 1/(1-p) - p^(n+1)/(1-p) but if I take the limit of the second term, wouldn't it diverge?

Not in the p-adic numbers it doesn't diverge. What does p^(n+1) converge to?

 

[arturo_026]

Not in the p-adic numbers it doesn't diverge. What does p^(n+1) converge to?

Ohhh! in the p-adic metric as n --> inf p^(n+1) goes to zero since p-adic abs. value of p^(n+1) = p^-(n+1)