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This old question recently popped up, and I thought about how I would answer it. My approach was nothing like the approaches in the given answers, which makes me wonder, Is my approach invalid in some way?

Edited to account for objection raised in answer by Lukas Heger

I reasoned that a product of a finite number of finite integers cannot become infinite. Thus, if $n=\prod p_i^{a_i}$, then $n \rightarrow \infty$ only if one or more of $p_i \rightarrow \infty$, or if $i \rightarrow \infty$, or if one or more of $a_i \rightarrow \infty$ (or some combination of those). These conditions are sufficient to ensure that the product representing $n$ will have either an infinite number of prime factors (not necessarily distinct) or have prime factors that are arbitrarily large.

In any case it will be true that $\phi(n)=\prod (p_i-1)p_i^{a_i-1} \rightarrow \infty$ because necessarily one or more of $(p_i-1) \rightarrow \infty$ or $i \rightarrow \infty$ or one or more of $(a_i-1) \rightarrow \infty$

Am I right, or have I missed something?

Keith Backman
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1 Answers1

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I think what you need is the following fact:

Lemma: Let $f(n)$ be a positive multiplicative function. If $f(q)\to+\infty$ as prime power $q\to\infty$. Then $f(n)\to+\infty$ as positive integer $n\to+\infty$.

Suppose this lemma is true, we can set $f(n)=\varphi(n)$ so it follows from $\varphi(p^a)=p^a(1-1/p)>\frac12p^a$ that $\varphi(n)/n^{1-\delta}\to+\infty$ as $n\to+\infty$ for any fixed $\delta>0$.

Proof. By definition, for all $K>1$ there exists a prime power $q_0$ such that for all prime power $q>q_0$ there is $f(q)>K$.

Now, let $Q$ be all the prime divisors of a large positive integer $n$. Then we can partition set $Q$ into three subsets:

$$ Q_1=\{q|n:q>q_0\},\quad Q_2=\{q|n:q\le q_0,f(q)\ge1\},\quad Q_3=\{q|n:q\le q_0,f(q)<1\}. $$

As a result, we have

$$ f(n)=\prod_{q_1\in Q_1}f(q_1)\prod_{q_2\in Q_2}f(q_2)\prod_{q_3\in Q_3}f(q_3)>K\underbrace{\prod_{q_3\in Q_3}f(q_3)}_{\text{bounded below}}, $$

in which the right most sum is bounded below because $Q_3$ is finite and $f(n)$ is positive. Thus, we see that $f(n)\to+\infty$ as $n\to+\infty$.

TravorLZH
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