**Note:** I am reposting this [question](https://math.stackexchange.com/q/4436917/369800) 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](https://www.hindawi.com/journals/ijsa/2006/065023/)". 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.



Inequality with slowly varying functions
$\newcommand{\al}{\alpha}$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 $x>e$.


