P-adic Integer has Unique Coherent Sequence Representative/Lemma 3

Theorem
Let $p$ be a prime number.

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rational numbers $\Q$.

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

Let $\sequence{\beta_n}$ be a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$ such that:
 * $\forall j \in \N : \exists \mathop {\map N j} \ge j : \forall m, n \in \N: m, n \ge \map N j : \norm {\beta_n - \beta_m} \le p^{-\paren {j + 1} }$

Let $\sequence{\alpha_n}$ be a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$ such that:
 * $\forall j \in \N: \norm {\alpha_j - \beta_{\map N j} }_p \le p^{-\paren {j + 1} }$

Then:
 * $\sequence {\alpha_n}$ and $\sequence {\beta_n}$ are representatives of the same $p$-adic number in $\Q_p$.

Proof
From Representatives of same P-adic Number iff Difference is Null Sequence, it needs only to be shown that $\sequence {\alpha_j - \beta_j}$ is a null sequence.

Let $\epsilon \in \R_{> 0}$.

From Sequence of Powers of Number less than One, the sequence $\sequence {p^n}$ is a null sequence.

Then there exists $j \in \N$ such that $p^{-j} < \epsilon$.

Then for all $i \ge \map N j$:

It follows that $\sequence {\alpha_n - \beta_n}$ is a null sequence by definition.