How to prove the tail behavior of the sum of random variables with one dominating? Assume I have given independent, continous random variables $X_1, \ldots, X_n$ and assume that they all have support $[-\infty, \infty]$. If $X_1$ asymptotically dominates all others, i.e.
$$f_{X_i}(\epsilon \cdot x)/f_{X_1}(x) \to 0 \quad\text{for }x \to \infty$$
for all $i \in 2,...,n$ and all fixed $\epsilon>0$, then, in my opinion, it should also hold that
$$f_{X_1+\ldots, + X_n}(x) \sim f_{X_1}(x) \quad \text{for }x\to \infty$$
But how would I generally prove that? It reminds me of something from regularly varying random variables, where if $X$ and $Y$ are regularly varying, then
$$f_{X+Y}(x)/(f_X(x)+f_Y(x))\to 1,$$
but I cannot quite figure out use the proof of the regular variation in this problem

Edit: After some useful remarks, I do not think that my assertion holds so let's restart;
Assume that I have two continuous, independent random variables $X$ and $Y$ with unlimited support $[-\infty,\infty]$. Assume the survival function of $X$ dominates $Y$, i.e.
$$\bar F_Y(\epsilon x)/\bar F_X(x) \to 0 \text{ for }x \to \infty,$$
where $bar F_X(x)=\mathbb P(X>x)$ is the survival function for any constant $\epsilon>0$ . Does it then also hold that
$$\bar F_{X+Y}/\bar F_X(x) \to 1 \text{ for }x \to \infty$$
 A: By "$f_{X_1+\ldots, + X_n}(x) \sim f_{X_1}(x) \quad \text{for }x\to \infty$", do you mean that the pdf of $\sum X_i$ minus the pdf of $X_1$ converges to 0? I don't think that's always true. I'm imagining a counterexample where $f_{X_1}$ has an infinite number of increasingly narrow peaks where the pdf takes value 1, separated by troughs. Suppose that $f_{X_1}(n) = 1$ whenever $n$ is an integer. Suppose also that $X_2$ is uniformly distributed on $[0, 1]$. Then adding $X_2$ has the effect of smearing out the peaks of $f_{X_1}$, so for sufficiently large $x$, $f_{X_1 + X_2}$ doesn't get close to 1, so it can't converge to $f_{X_1}$ whose limit supremum is 1.
(In this case $f_{X_1}$ isn't a regularly varying function, but the question didn't explicitly state the condition that it should be.)
Edit: @whuber points out that $X_2$ was supposed to have unbounded support. I believe the counterexample still holds if $X_2$ follows e.g. a standard normal distribution.

A counterexample for the new version of the question: suppose $X$ has a Laplace distribution with pdf $f_X(x) = \frac{1}{2}e^{-|x|}$, and that $Y \sim N(2, 1)$. Let $Z = X + Y$.
Then for $x > 1$,
$$
P(X > x) = \frac{1}{2}e^{-x}
$$
The distribution of $Y$ is symmetric about 2, and for $x>1$ and any $t$, $P(X > x-1-t)+P(X>x-1+t)\ge 2P(X>x)$, so we have:
$$
\begin{align}
P(Z>x \cap Y \notin [1,3]) &= \int_1^\infty f_Y(1+t)P(X>x-1-t) + f_Y(1-t)P(X>x-1+t)dt \\
&\ge \int_1^\infty 2f_Y(1+t)P(X>x)dt \\
&=P(X>x \cap Y \notin [1,3]) \\
&=P(X>x)P(Y \notin [1,3])
\end{align}
$$
The contribution from the case $Y \in [1,3] $ is easy to bound:
$$
P(Z>x|Y\in [1,3]) \ge P(Z>x|Y=1) = P(X\ge x-1) = \frac{1}{2}e^{1-x} \\
$$
Combining these:
$$
\begin{align}
P(Z > x ) &= P(Z > x | Y \in [1,3])P(Y \in [1,3]) + P(Z>x|Y\notin [1,3)P(Y\notin [1,3]) \\
&\ge P(X > x-1)P(Y \in [1,3]) + P(X>x)P(Y\notin [1,3]) \\
&= \frac{1}{2}e^{1-x}P(Y \in[1,3]) + \frac{1}{2}e^{-x}P(Y\notin [1,3]) \\
&=P(X>x)(eP(Y\in [1,3]) + P(Y\notin[1,3]))
\end{align}
$$
This gives an asymptotic lower bound for $\bar{F}_{X+Y}/\bar F_X(x)$ that's greater than 1.
