TITLE: Asymptotic of series
QUESTION [6 upvotes]: Let $x_n$ be a sequence of positive numbers with the property that
$$\sum_n \frac{x_n}{n \log(n)^3}< \infty.$$
Can we show that
$$\limsup_{\varepsilon \downarrow 0}\frac{\sum_{i=1}^{\infty} \exp(-\varepsilon x_n)}{\sum_{i=1}^{\infty} \exp(-\varepsilon \log(i)^2)}>0?$$

REPLY [3 votes]: This isn't a complete solution, but it shows that the limit is $+\infty$ for "most" $t$ (the meaning will be made precise near the end -- here my $t$ is your $\varepsilon$). The proof shows that the "large" terms of $x_n$ can be chosen so that the sum of the $e^{-t(\ln{n})^2}$ (with $n$ running over the indices of the large terms) can be (for some $t$, relatively few) the most significant part of the sum $\sum_{n \geq 1}{e^{-t(\ln{n})^2}}$.
First, it's important to get a good idea of how the series $F(t)=\sum_{n \geq 1}{e^{-t(\ln{n})^2}}$ behaves for small $t > 0$. We'll be very crude in the estimate, but the result will be more than sufficient.
By series-integral comparison, it's easy to see that $|F(t)-F_i(t)| \leq 2$, where $F_i(t) = \int_0^{\infty}{e^{-t(\ln{u})^2}du}$, because the part from $0$ to $1$ is bounded in modulus by $1$, and the general term is a (smooth) decreasing function of $n$ bounded by $1$.
Writing $u = e^v$, we see that $F_i(t) = \int_{-\infty}^{\infty}{e^{v-tv^2}dv} = \frac{\sqrt{\pi}}{\sqrt{t}}\exp\left(\frac{1}{4t}\right)$ by completing the square.
We can obviously assume $x_1=x_2=0$ (that's not very important but it makes the computations simpler).
So, let now $S = \{n \geq 3,\, 2x_n \geq \ln^2{n}\}$: we know that $\sum_{n \in S}{\frac{1}{2n\ln{n}}} < \infty$.
For each $p \geq 1$, let $N_p = \{n \in S,\, e^p < n \leq e^{p+1}\}$. That the previous series converges can be rewritten as $\sum_{p \geq 1}{\frac{N_p}{pe^p}} < \infty$. Write $u_p = N_pe^{-p}$: it is a bounded nonnegative sequence (by $e-1$) such that $\sum_{p \geq 1}{p^{-1}u_p} < \infty$.
In particular, let $T$ be the set of indices $p$ such that $u_p > 1/10$, and let $b_n$ be an increasing sequence of image $T$. We know that $\sum_{n \geq 1}{\frac{1}{b_n}} \leq \sum_{p \in T}{\frac{10u_p}{p}} < \infty$. By a standard analysis lemma, it follows that $n = o(b_n)$.
Then $$\sum_{n \in S}{e^{-t(\ln{n})^2}} = \sum_{p \geq 1}{\sum_{n \in S \cap (e^p,e^{p+1}]}{e^{-t(\ln{n})^2}}} \leq \sum_{p \geq 1}{u_pe^{p-tp^2}} \leq \frac{1}{10}\sum_{p \geq 1}{e^{p-tp^2}}+(e-1)\sum_{n \geq 1}{e^{b_n-tb_n^2}}.$$
So, define $G(t) = \sum_{n \geq 1}{e^{n-tn^2}}$ for each $t > 0$. We want to show an explicit inequality between $G(t)$ and $t^{-1/2}e^{1/4t}$.
As $x \longmapsto x-tx^2$ is increasing for $0 \leq x \leq 0.5/t$, and decreasing for larger $x$, and as this function is bounded by $1/4t$, it follows that $$0 \leq G(t) \leq 2e^{0.25/t}+\int_0^{\infty}{e^{x-tx^2}dx} \leq 2e^{0.25/t}+e^{.25/t}\int_{\mathbb{R}}{e^{-t(x-1/2t)^2}dx} \leq \left(2+\frac{\sqrt{\pi}}{\sqrt{t}}\right)\exp{\frac{1}{4t}}.$$
Let's study $H(t)=\sum_{n\geq 1}{e^{b_n-tb_n^2}}$ -- we want to show that this quantity is, for small enough $t$, bounded by some $B(1+o(1))\frac{e^{0.25t^{-1}}}{t^{1/2}}$, where $B$ is a constant.
First, let's assume this is the case. Then, for small enough $t$, we thus have $\sum_{n \in S}{e^{-t(\ln{n})^2}} \leq \frac{1}{\sqrt{t}}\exp\left(\frac{1}{4t}\right)(c+o(1))$, where $c < \sqrt{\pi}$ (here $c = B(e-1)+\frac{\pi^{1/2}}{10}$). It follows that there exists some $k > 1$ such that for small enough $t >0$, $F(t) \geq k\sum_{n \in S}{e^{-t(\ln{n})^2}}$, and thus that $\sum_{n \notin S}{e^{-t(\ln{n})^2}} = \Theta(F(t))$. From this, it follows that for small enough $t > 0$,
$\frac{\sum_{n \geq 1}{e^{-tx_n}}}{\sum_{n\geq 1}{e^{-t(\ln{n})^2}}} \geq \Theta\left(\frac{F(t/2)}{F(t)}\right) = \Theta\left(\frac{F_i(t/2)}{F_i(t)}\right) = \Theta(e^{1/4t})$, which obviously concludes.
Now, we still have to compare $H(t)$ to $G(t)$. We split the sum in three:
$\sum_{1 \leq p \leq 0.5t^{-1}-t^{-1/2}}{e^{p-tp^2}} \leq e^{1/4t}+\int_{-\infty}^{1/2t-1/\sqrt{t}}{e^{x-tx^2}\,dx} = e^{1/4t}+e^{1/4t}\int_{-\infty}{-1/\sqrt{t}}{e^{-tu^2}dx} \leq e^{1/4t}+e^{1/4t}\times \frac{0.14}{\sqrt{t}}$.
Similarly, $\sum_{p \geq 0.5/t+t^{-1/2}}{e^{p-tp^2}} \leq e^{1/4t}+e^{1/4t}\int_{1/\sqrt{t}}^{\infty}{e^{-tx^2}dx} <e^{1/4t}(1+0.14t^{-1/2})$.
(Note that, if we had chosen $t^{-2/3}$ instead of $t^{-1/2}$, then the two sums would have been negligible before $\frac{e^{1/4t}}{\sqrt{t}}$ -- so if $T$ contains the whole interval $[1/2t-t^{-2/3},1/2t+t^{2/3}]$, then $H(t)$ is very large and we won't be able to say anything. Fortunately, as explained below, this is very rare).
Thus, if $T_t$ is the cardinality of $T \cap [1/2t-1/\sqrt{t},1/2t+1/\sqrt{t}]$, then $H(t) < (2+T_t+0.28/\sqrt{t})e^{1/4t}$. So if $|T_t| \leq 0.648t^{-1/2}$ (the $0.648$ comes from the formulas for $B$ and $c$ above), we are done.
Except that there's no guarantee that this happens for all $t$ small enough:  if $T = \bigcup_{n \geq 5}{[2^{2n-1}-3^n,2^{2n-1}+3^n]}$, then $|T_{4^{-n}}| = 2^{n+1}+1 > 1.49 \sqrt{4^n}$.
However, what we can say is that, if $0 < t_0 < 1/2$,
$\int_{t_0}^{1/2}{T_t\sqrt{t}\frac{dt}{t}} = 2\sum_{n \in T}{\int_{\sqrt{2}}^{\sqrt{1/t_0}}{1(2n \in [u^2-2u,u^2+2u])\frac{du}{u^2}}}$. Solving the quadratic equations -- since $u^2 = \Theta(n)$ when $u^2-u \leq n \leq u^2+u$ -- yields that $$\int_{t_0}^{1/2}{T_t\sqrt{t}\frac{dt}{t}} \leq C\sum_{n \in T}{n^{-1}} < \infty.$$
In other words: there exists a set $U \subset (2,\infty)$ (here $U$ is the set of $t$ such that $|T_{2/t}|/\sqrt{t/2}$ or $|T_{1/t}|/\sqrt{t}$ is greater than the $0.648$ from above) such that $\int_U{\frac{dt}{t}} < \infty$ and such that if $t_n \rightarrow 0$ and $1/t_n \notin U$, then $\frac{1}{F(t_n)}\sum_{q \geq 1}{e^{-t_n(\ln{q})^2}} \rightarrow \infty$.
This, of course, doesn't settle the problem. There is the question of these "exceptional" $t$. It's surprising that this arises because there's no obvious reason why the behavior should be so "erratic" when $t$ changes.
Note that I haven't used the full power of the original assumption (I think the second one doesn't matter -- we can just make a few terms huge, say $a_n = n(\ln{n})^{5/2}$, at $n=2^{k^k}$) about the convergent series. I just used that most terms would be smaller than $(\ln{n})^2/2$. I'm not sure if this is helpful, because it could be that there are still too many "large" $x_n$ for the contribution of the $e^{-tx_n}$ with $x_n$ "small" to matter (at least for some $t$), or that the $x_n$ aren't small enough to make a difference.