Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 7
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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 |
$\blacksquare$