# Proof that an archimedean absolute value on $\mathbb Q$ is equivalent to $|\ |_\infty$

I have included a bolded comment in a step in the part of Gouvea's proof of Ostrowski's theorem where he shows that an archimedean absolute value on $\mathbb Q$ is equivalent to $|\ |_\infty$ (the regular absolute value). I will write my question about the comment below. Here is the relevant excerpt of the proof:

Suppose, first, that $|\ |$ is archimedean. We want to show, in this case, that it is equivalent to the $\infty-$adic absolute value. Since $|\ |$ is archimedean, $\exists n\in\mathbb Z: |n|>1$. Then $|-n| = |n| > 1$, so suppose WLOG that $n \geq 0$. In fact, we know that $n \geq 2$. Let $n_0$ be the least such positive integer $n$. Then since $n_0 > 1$ and $|n_0| > 1$, we can pick $\alpha\in\mathbb R$ such that $$|n_0| = n_0^\alpha.$$ We claim that $\forall x\in \mathbb Q$, $|x| = |x|_\infty^\alpha$. Note that this already holds for $x = 0$ (and $x = 1$), and that if it holds for positive integers, then it holds for positive rational numbers, and then it holds for negative rational numbers, so it holds for all of $\mathbb Q$. Thus, it is sufficient to demonstrate this equality for positive integers. We will achieve this by showing that for every positive integer $n$, $|n| = n^\alpha$.

We know that the equality holds for $n = n_0$. Now take an arbitrary positive integer $n$, and write it "in base $n_0$," i.e., in the form $$n = a_0 + a_1n_0 + a_2n_0^2 + \cdots + a_kn_0^k,$$ with $k\geq 0$, $0\leq a_i < n_0$, and $a_k\neq 0$. Now taking absolute values, we get \begin{align*} |n| & = |a_0 + a_1n_0 + a_2 n_0^2 + \cdots + a_kn_0^k| \\ & \leq |a_0| + |a_1| |n_0| + |a_2| |n_0|^2 + \cdots + |a_k| |n_0|^k \\ & = |a_0| + |a_1| n_0^\alpha + |a_2| n_0^{2\alpha} + \cdots + |a_k| n_0^{k\alpha}. \end{align*} Since we chose $n_0$ to be the smallest positive integer whose absolute value was greater than $1$, we know that $|a_i| \leq 1$, so that we get \begin{align*} |n| & \leq |a_0| + |a_1| n_0^\alpha + |a_2| n_0^{2\alpha} + \cdots + |a_k| n_0^{k\alpha} \\ & \leq 1 + n_0^\alpha + n_0^{2\alpha} + \cdots + n_0^{k\alpha} \\ & = n_0^{k\alpha}(1 + n_0^{-\alpha} + n_0^{-2\alpha} + \cdots + n_0^{-k\alpha} ) \\ & = n_0^{k\alpha} \sum_{i=0}^k n_0^{-i\alpha} \\ & \leq n_0^{k\alpha} \sum_{i=0}^\infty n_0^{-i\alpha} \textbf{ (why is this not strict?)} \\ & = n_0^{k\alpha} \frac{1}{1 - n_0^{-\alpha}} = n_0^{k\alpha} \frac{n_0^\alpha}{n_0^\alpha - 1}. \end{align*} Since $n_0^\alpha > 1$, we have that $C:= n_0^\alpha/(n_0^\alpha - 1) > 0$, and $$|n| \leq C n_0^{k\alpha} \leq Cn^\alpha.$$ To see why we were able to tack on $Cn^\alpha$, just look at the expansion of $n$ "in base $n_0$" to see that $n_0^k \leq n$. The above formula applies for every positive integer $n$, and thus letting $n$ and $m$ be any positive integers, we can apply the formula to $n^m$, \begin{align*} |n^m| & \leq C n^{m\alpha} \\ |n| & \leq \sqrt[m]{C} n^\alpha . \end{align*} If we leave $n$ fixed and let $m$ become arbitrarily large, then since $C>0$ is fixed, we see that $\sqrt[m]{C}$ becomes arbitrarily close to $1$, and thus the above inequality implies that $|n| \leq n^\alpha$.

In the second part of this proof, Gouvea goes on to show that $|n| \geq n^\alpha$, and then he is able to conclude that $|n| = n^\alpha$. Thus, it is essential that in the excerpt above, we find that $|n| \leq n^\alpha$. However, since the terms of the sum $\sum_{i=0}^\infty n_0^{-i\alpha}$ are each nonnegative, I think it must be that $\sum_{i=0}^k n_0^{-i\alpha} < \sum_{i=0}^\infty n_0^{-i\alpha}$, and thus $$n_0^{k\alpha} \sum_{i=0}^k n_0^{-i\alpha} < n_0^{k\alpha} \sum_{i=0}^\infty n_0^{-i\alpha},$$ contradicting the step in Gouvea's proof. Is my logic or my reading of his proof wrong?

I just realized that since I can say $C > 1$, the proof still works even with a strict inequality: we end up with \begin{align*} |n^m| & < C n^{m\alpha} \\ |n| & < \sqrt[m]{C} n^\alpha . \end{align*} If we leave $n$ fixed and let $m$ become arbitrarily large, then since $C>1$ is fixed, we see that $\sqrt[m]{C}$ becomes arbitrarily close to $1$ from above, and thus the above inequality implies that $|n| \leq n^\alpha$.