# Leigh.Samphier/Sandbox/Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 7

## Theorem

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.

Then:

$1 + p^k + p^{2k} + p^{3k} + \ldots = \dfrac 1 {1 - p^k}$

## Proof

Let $S_n$ be the partial sum:

$\ds S_n = \sum_{j = 0}^n p^{j k}$

We have:

 $\ds \paren{1 - p^k} S_n$ $=$ $\ds \paren{1 - p^k} \sum_{j = 0}^n p^{j k}$ $\ds$ $=$ $\ds \paren{\sum_{j = 0}^n p^{j k} } - p^k \paren{\sum_{j = 0}^n p^{j k} }$ Distributing the partial sum across $1 - p^k$ $\ds$ $=$ $\ds \paren{\sum_{j = 0}^n p^{j k} } - \paren{\sum_{j = 0}^n p^{\paren{j + 1} k} }$ Distributing $p^k$ across the partial sum $\ds$ $=$ $\ds 1 - p^k + p^k - p^{2 k} + p^{2 k} - \ldots - p^{n k } + p^{n k } - p^{\paren{n + 1} k}$ Re-arranging terms $\ds$ $=$ $\ds 1 - p^{\paren{n + 1} k}$ Removing cancelling terms

 $\ds \leadsto \ \$ $\ds \norm{\paren{1 - p^k} S_n - 1}_p$ $=$ $\ds \norm{1 - p^{\paren{n + 1} k} - 1}_p$ $\ds$ $=$ $\ds \norm{- p^{\paren{n + 1} k} }_p$ Remove cancelling terms $\ds$ $=$ $\ds \norm{p^{\paren{n + 1} k} }_p$ Norm of Negative in Division Ring $\ds$ $=$ $\ds \dfrac 1 {p^{\paren{n + 1} k} }$ Definition of P-adic Norm on Rational Numbers $\ds$ $\to$ $\ds 0$ as $n \to \infty$

 $\ds \leadsto \ \$ $\ds \lim_{n \mathop \to \infty} \paren{1 - p^k} S_n$ $=$ $\ds 1$ Definition of Convergent P-adic Sequence $\ds \leadsto \ \$ $\ds \lim_{n \mathop \to \infty} S_n$ $=$ $\ds \dfrac 1 {\paren{1 - p^k} }$ Multiple Rule for Sequences in Normed Division Ring $\ds \leadsto \ \$ $\ds \sum_{j = 0}^\infty p^{j k}$ $=$ $\ds \dfrac 1 {\paren{1 - p^k} }$ Definition of Series

$\blacksquare$