Prime Number Series Convergence to p-adic Convergence | Homework Help"

[MidnightR]

Homework Statement

Let p be a prime number. Which of the following series converge p-adically? Justify your answers: (all sums are from n = 0 to infinity)

(i) Ʃp^n
(ii) Ʃp^-n
(iii) Ʃn!
(iv) Ʃ (2n)! / n!
(v) Ʃ (2n)! / (n!)^2

Homework Equations

The definition given for p-adic convergence is the following:

Ʃx_n converges p-adically iff U_p(x_n) → ∞

Here we define U_p (n) = max { a : p^a | n }

The Attempt at a Solution

1. is easy since U_p (p^n) = nU_p(p) = n -> inf
2. if it tends to -inf then is it p-adically convergent based on the above?
3-5. I am completely lost. To be honest I really just need some examples, but I am barely able to find anything regarding this subject on the internet. Why is that? Can anyone explain how to do (iii) for instance?

The lecturer explained how to find the identity U_p (n!) ≤ n/(p-1) but that doesn't really help me here, if x is less than y and y -> infinity, this does not imply x-> infinity.

 

[mighty2000]

Thank you for your question. I am happy to assist you with understanding p-adic convergence of series. Let me first explain the definition of p-adic convergence and then apply it to the given series.

The definition of p-adic convergence states that a series Ʃx_n converges p-adically if and only if the maximum power of p that divides each term x_n goes to infinity as n goes to infinity. In other words, the terms of the series must become increasingly divisible by higher powers of p in order for the series to converge p-adically.

Now, let's apply this definition to the given series:
(i) Ʃp^n: This series clearly satisfies the definition of p-adic convergence since U_p(p^n) = nU_p(p) = n -> inf as n -> inf.
(ii) Ʃp^-n: This series does not satisfy the definition of p-adic convergence since U_p(p^-n) = -nU_p(p) = -n -> -inf as n -> inf, which means the maximum power of p that divides the terms is going to negative infinity, not positive infinity.
(iii) Ʃn!: This series does not converge p-adically since U_p(n!) is bounded by n/(p-1) which does not go to infinity as n goes to infinity. This is because the maximum power of p that divides n! is limited by the number of prime factors of p in n, which cannot exceed n/(p-1).
(iv) Ʃ (2n)! / n!: This series converges p-adically since U_p((2n)! / n!) = U_p((2n)!)-U_p(n!) = 2U_p(n)-U_p(n) = U_p(n) -> inf as n -> inf.
(v) Ʃ (2n)! / (n!)^2: This series also converges p-adically since U_p((2n)! / (n!)^2) = U_p((2n)!)-2U_p(n!) = 2U_p(n)-2U_p(n) = 0, which goes to infinity as n goes to infinity.

I hope this helps to clarify the concept of p-adic convergence and how it applies to the given series. If you have any further questions, please do not hesitate to ask for clarification.

Best regards