I am trying to wrap my head around two theorems of Diophantine approximation: Khinchin's theorem and the Duffin and Schaeffer conjecture. To the best of my understanding, here is what they say:
Khinchin's theorem. Let $\psi$ be a function $\mathbb{N} \to \mathbb{R}^+$ such that $q\psi(q)$ is non-increasing. Call $\alpha$ "$\psi$-approximable" if there exist infinitely many $\frac{p}{q}$ such that $|\alpha - \frac{p}{q}| < \frac{\psi(q)}{q}$. (Let's call this inequality (⋆).)
If $\sum_{q = 1}^\infty \psi(q)$ converges, then almost every $\alpha$ is not $\psi$-approximable.
If $\sum_{q = 1}^\infty \psi(q)$ diverges, then almost every $\alpha$ is $\psi$-approximable.
Duffin and Schaeffer conjecture. Let $\psi$ be any function $\mathbb{N} \to \mathbb{R}_{\geq 0}$. Call $\alpha$ "$\psi_r$-approximable" ($r$ for reduced) if there exist infinitely many reduced $\frac{p}{q}$ such that (⋆) holds.
If $\sum_{q = 1}^\infty \frac{\phi(q)\psi(q)}{q}$ converges, then almost every $\alpha$ is not $\psi_r$-approximable.
$\sum_{q = 1}^\infty \frac{\phi(q)\psi(q)}{q}$ diverges, then almost every $\alpha$ is $\psi_r$-approximable.
(Note that $\phi(q)$ is Euler's totient function.) Now, suppose that $\psi$ is a function satisfying the conditions of Khinchin's theorem. In this case, an irrational number $\alpha$ is $\psi$-approximable if and only if $\alpha$ is $\psi_r$-approximable. Here's the reason. In this case, $\psi$ is decreasing, so if $\frac{km}{kn}$ satisfies (⋆) then so does $\frac{m}{n}$. And if $\frac{m}{n}$ satisfies (⋆), then for $k$ large enough $\frac{km}{kn}$ will not satisfy (⋆). So the only way infinitely many fractions can satisfy (⋆) is if infinitely many reduced fractions do.
Here is what this seems to imply: for such functions $\psi$, the series $\sum_{q = 1}^\infty \psi(q)$ converges if and only if the series $\sum_{q = 1}^\infty \frac{\phi(q)\psi(q)}{q}$ converges. Indeed, $\sum_{q = 1}^\infty \psi(q)$ converges $\iff$ a.e. $\alpha$ is not $\psi$-approximable $\iff$ a.e. $\alpha$ is not $\psi_r$-approximable $\iff$ $\sum_{q = 1}^\infty \frac{\phi(q)\psi(q)}{q}$ converges.
Here is my question. Is it really true that, if $\psi: \mathbb{N} \to \mathbb{R}^+$ is a function with the property that $q\psi(q)$ is non-increasing, then $\sum_{q = 1}^\infty \psi(q)$ converges $\iff $ $\sum_{q = 1}^\infty \frac{\phi(q)\psi(q)}{q}$ converges? Is there some "easy" way to see why?
One direction is obvious, since $\psi(q)$ is always larger than $\frac{\phi(q)\psi(q)}{q}$. But the other direction seems interesting and surprising. For example, if we plug in $\psi(q) = \frac{1}{q\log q\log\log q}$, then we have that $\sum_{q = 2}^\infty \frac{1}{q\log q\log\log q}$ diverges, so the above would say that $\sum_{q = 2}^\infty \frac{\phi(q)}{q^2\log q\log\log q}$ diverges, too. (I've started at $q = 2$ to avoid dividing by 0.)
Thanks for your help!
Note. I have no advanced knowledge of either number theory or analysis. So if you could make your answer(s) user-friendly, I'd appreciate it. :-)
Edit: $\sum_{q = 1}^\infty \psi(q)$ converges $\iff$ a.e. $\alpha$ is not $\psi$-approximable $\iff$ a.e. $\alpha$ is not $\psi_r$-approximable $\iff$ $\sum_{q = 1}^\infty \frac{\phi(q)\psi(q)}{q}$ converges. I didn't have those "not"s before.
Edit 2: Changed to say "In this case, an irrational number $\alpha$ is $\psi$-approximable if and only if $\alpha$ is $\psi_r$-approximable." If $\alpha$ is rational, $\psi$-approximability and $\psi_r$-approximability are not equivalent.
The answer to this question is yes! Proposition 4.7 of these lecture notes states the following:
There exists $c > 0$ such that for any positive non-increasing function $\psi$ one has
$$\forall N \in \mathbb{N}, \hspace{1em}\sum_{q = 1}^N\frac{\phi(q)}{q}\psi(q) > c\sum_{q = 1}^N\psi(q).$$
Obviously $$\sum_{q = 1}^N\psi(q) > \sum_{q = 1}^N\frac{\phi(q)}{q}\psi(q),$$
so the convergence or divergence of these sums as $N \to \infty$ is equivalent.
Note that the condition placed on $\psi$ here is weaker than what I wrote above: $\psi(q)$, not necessarily $q\psi(q)$, is assumed to be non-increasing. In fact, the author states that this condition can be further weakened: it suffices for the function $\displaystyle\frac{\psi(q)}{q^v}$ to be non-increasing for some $v \geq 0$.
Such a fascinating, surprising result!