User:Leigh.Samphier/Sandbox/Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 9

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 $\R$:

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.