Partial Sums of P-adic Expansion forms Coherent Sequence

Theorem
Let $p$ be a prime number.

Let $\ds \sum_{n \mathop = 0}^\infty d_n p^n$ be a $p$-adic expansion.

Let $\sequence{\alpha_n}$ be the sequence of partial sums; that is:
 * $\forall n \in \N :\alpha_n = \ds \sum_{i \mathop = 0}^n d_i p^i$.

Then $\sequence{\alpha_n}$ is a coherent sequence.

Proof
From the definition of a coherent sequence, it needs to be shown that $\sequence{\alpha_n}$ is a sequence of integers such that:
 * $(1): \quad \forall n \in \N: 0 \le \alpha_n < p^{n + 1}$
 * $(2): \quad \forall n \in \N: \alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1}}$

That the sequence $\sequence{\alpha_n}$ is a sequence of integers follows immediately from the assumption that the series begins at $n = 0$ and so the terms of each summation are integers.

First note:

Partial Sums satisfy (1)
This is proved by induction on $n$:

Basis for the Induction
$n = 0$

By definition: $\alpha_0 = d_0$

By definition of a $p$-adic expansion:
 * $0 \le d_0 < p$

So shown for basis for the induction.

Induction Hypothesis
This is our induction hypothesis:
 * $0 \le \alpha_k < p^{k + 1}$

Now we need to show true for $n=k+1$:
 * $0 \le \alpha_{k + 1} < p^{k + 2}$

Induction Step
This is our induction step:

Since:
 * $\alpha_{k + 1} = \alpha_k + d_{k + 1} p^{k + 1}$

then:
 * $0 \le \alpha_{k + 1} < p^{k + 2}$

By induction:
 * $\forall n \in \N: 0 \le \alpha_n < p^{n + 1}$

Partial Sums satisfy (2)
Let $n \in \N$. Since:
 * $\alpha_{n + 1} - \alpha_n = d_{n + 1}p^{n + 1}$

Then:
 * $p^{n + 1} \divides \alpha_{n+1} - \alpha_n$

So:
 * $\alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1} }$

The result follows.

Note
The theorem is stated for the special set of $p$-adic expansions where the series index begins at $0$ and not the more general $m \in \Z_{\le 0}$.

From P-adic Integer has Unique P-adic Expansion Representative, it is seen that this set of $p$-adic expansions is indeed the $p$-adic integers.

Also see

 * Coherent Sequence is Partial Sum of P-adic Expansion