Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 9

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Theorem

Let $p$ be a prime number.

Let $a \in \Z, b \in Z_{> 0}$


Then:

$\ds \lim_{n \mathop \to \infty} \dfrac {a - \paren{p^{n+1} - 1} b } {p^{n+1}} = -b$


Proof

From Sequence of Reciprocals is Null Sequence:

$\ds \lim_{n \mathop \to \infty} \dfrac 1 n = 0$

From Combined Sum Rule for Real Sequences:

$\ds \lim_{n \mathop \to \infty} \dfrac {n - 1} n = \lim_{n \mathop \to \infty} 1 - \dfrac 1 n = 1$

From Limit of Subsequence equals Limit of Real Sequence:

$\ds \lim_{n \mathop \to \infty} \dfrac {p^{n + 1} - 1} {p^{n + 1} } = 1$


Lemma 8

In the real numbers $\R$:

$\ds \lim_{n \mathop \to \infty} \dfrac a {p^{n+1}} = 0$

$\Box$


From Combined Sum Rule for Real Sequences:

$\ds \lim_{n \mathop \to \infty} \dfrac a {p^{n + 1} } - b \paren {\dfrac {p^{n + 1} - 1} {p^{n + 1} } } = -b$


The result follows.

$\blacksquare$