Inequality with slowly varying functions Note: I am reposting this question from Math Stack Exchange, which failed to receive an answer after several weeks and a bounty. Also, I believe it fits the requirements for this website, as it relates to a research paper.
Question
Let $X$ be a random variable with distribution function $F$ on a probability space $(\Omega, \mathcal F, P)$.
Suppose that there exist $\alpha \in (0,2)$ and a slowly varying function $\ell(\cdot)$ such that
$$
\bar F(x) := 1 - F(x) = \frac{C_1(x)}{x^\alpha} \ell(x) \quad \text{and} \quad F(-x) = \frac{C_2(x)}{x^\alpha} \ell(x) \quad \text{for $x > 0$,}
$$
where $C_1(\cdot), C_2(\cdot)$ are non-negative functions with $C_i := \lim_{x \to \infty} C_i(x)$, and $C_1 + C_2 > 0$.
Why then does the following hold?

There exist $C, \tilde C > 0$ such that
$$ \sum_{n=1}^\infty P\big( |X| > a_n \big) \leq \sum_{n=1}^\infty \frac{C}{nf(n)} \leq \tilde C \int_1^\infty \frac{dt}{t f(t)}, 
$$
where we define $a_n := [n f(n) \ell(n) ]^{1/\alpha}$ for an arbitrary positive function $f$ with the properties
$$ \limsup_{t \to \infty} \sup_{0 \leq t \leq x} \frac{f(t)}{f(x)} < \infty \quad \text{and} \quad \int_1^\infty \frac{dt}{tf(t)}< \infty.
$$

If not, what if we also required that $f$ be slowly varying, too?
Background and Thoughts
This comes from Cai's 2006 paper "Chover-Type Laws of the Iterated Logarithm for Weighted sums of $\rho^*$-Mixing Sequences". Cai writes on page 5 that this is "easily seen" based on the representation of $F$ above. I don't see why. What follows below is my attempt so far.
$$
\begin{aligned}
\sum_{n=1}^\infty P\big( |X| > a_n \big) &= \sum_{n=1}^\infty \Big[ P\big( X > a_n \big) + P\big( X < - a_n \big)  \Big] \\
&\leq \sum_{n=1}^\infty \Big[ \bar F(a_n) + F(-a_n)  \Big] \\
&= \sum_{n=1}^\infty \frac{C_1(a_n) + C_2(a_n)}{a_n^\alpha} \ell(a_n) \\
&\leq \sum_{n=1}^\infty \frac{C}{a_n^\alpha} \ell(a_n) = C \sum_{n=1}^\infty \frac{1}{nf(n)} \cdot \frac{\ell(a_n)}{\ell(n)},
\end{aligned}
$$
where the inequality on the last line holds for some $C>0$, since $C_1(\cdot)$ and $C_2(\cdot)$ are convergent.
If $\frac{\ell(a_n)}{\ell(n)} = \ell\Big( \big[ n f(n) \ell(n) \big]^{1/\alpha} \Big) \Big/ \ell(n)$ were bounded, then the first desired inequality would follow. However, it's not clear why this would have to hold.
Updates

*

*Indeed, I think the author implicitly uses the fact (?) that $\left\{ \frac{\ell(a_n)}{\ell(n)} \right\}$ is a bounded sequence several times in the paper. But I still don't know why that's the case.


*Although not stated in the paper, maybe we need some more restrictions on $f$, such that it is also a slowly varying function. If $f$ were slowly varying, then $u(x) := x^{1/\alpha}\cdot[f(x) \ell(x)]^{1/\alpha}$ would be regularly varying with coefficient $1/\alpha$. And $\frac{\ell(a_n)}{\ell(n)} = \frac{\ell\big(u(n)\big)}{\ell(n)}$. Since $u(x) \to \infty$ and is regularly varying, $\ell \circ u$ is slowly varying. Hence, $\frac{\ell(a_n)}{\ell(n)} = \frac{ \ell \big(u(n) \big)}{\ell(n)}$ is slowly varying. But, of course, slowly varying functions aren't necessarily bounded.
 A: $\newcommand{\al}{\alpha}$This inequality is false, for any $\al\in(0,2)$.
Indeed, consider first the case $\al\ne1$. Suppose that $C_1(x)+C_2(x)=1$ for all $x>1$,
\begin{equation}
    \ell(x)=e^{b\sqrt{\ln x}} \tag{1}\label{1}
\end{equation}
for some real $b$ and all $x>1$, and $f(t)=\ln^2 t$ for all $t>2$.
Then, reasoning as in your post, for all large enough $n$ we have
\begin{equation}
    P(|X|>a_n)\asymp\frac1{nf(n)}\frac{\ell(a_n)}{\ell(n)}
    =\frac1{n\ln^2 n}\frac{\ell(a_n)}{\ell(n)}, 
\end{equation}
$a_n=n^{1/\al+o(1)}$,
\begin{equation}
    \frac{\ell(a_n)}{\ell(n)}=\exp\{b(\sqrt{1/\al}-1+o(1))\sqrt{\ln n}\}. 
\end{equation}
Letting now $b=1$ if $\al\in(0,1)$ and $b=-1$ if $\al\in(1,2)$, we see that for all large enough $n$
\begin{equation}
    \frac{\ell(a_n)}{\ell(n)}>\ln n
\end{equation}
and hence
\begin{equation}
\sum_n  P(|X|>a_n)=\infty, 
\end{equation}
whereas $\int_1^\infty \frac{dt}{t f(t)}<\infty$, so that the inequality
\begin{equation}
    \sum_{n=1}^\infty P\big( |X| > a_n \big) \le \tilde C \int_1^\infty \frac{dt}{t f(t)}
\end{equation}
cannot hold for any real $\tilde C$.
In the case $\al=1$, instead of \eqref{1} similarly consider
\begin{equation}
    \ell(x)=\exp\frac{\ln x}{\ln\ln x}
\end{equation}
for $x>e$.
