Leigh.Samphier/Sandbox/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

$\blacksquare$