Does one of these conditions on a sequence imply the other one? Let ${(r_n)}_{n \geq 0}$ be a sequence of integers $\geq 2$. Set $q_n=\prod_{i=0}^{n-1} r_i$ (agreeing with $q_0=1$). I want to know whether one of these two conditions implies the other one (I think the answer is no but I don't find counter-examples). The first condition is 
$$(\Delta)\colon\quad \sum \frac{r_n \log r_n}{q_{n+1}} < \infty.$$
The second condition is 
$$(\Theta)\colon\quad \limsup_{\epsilon\to 0}\,\epsilon \sum_{i=0}^{k(\epsilon)}r_i\log r_i \to 0$$
where $k(\epsilon) = \max\{k \mid q_{k+1}^{-1} \geq \epsilon\}$.
For example:


*

*Any bounded sequence $(r_n)$ satisfies both conditions. This is obvious for $(\Delta)$, and not difficult for $(\Theta)$.

*The case $r_n=n+2$ satisfies both conditions. This is not difficult for $(\Delta)$, and this can be shown for $(\Theta)$ by using the inequalities $k! \geq k^{\frac{k}{2}}$ and $\sum_{i=1}^n k \log k \leq {(n \log n)}^2$. 
Incidentally, if none of these conditions implies the other one, I would be interested in a simplified formulation of condition $\Delta\cap\Theta$.

@Adayah's answer provides the proof of this lemma:
Lemma: Let ${(u_n)}_{n \geq 0}$ and ${(v_n)}_{n \geq 0}$ be two sequences of positive numbers such that $v_n \searrow 0$. If $\sum u_n v_n < \infty$ then $\epsilon \sum_{i=0}^{n(\epsilon)} u_i \to 0$ when $\epsilon \to 0^+$, where $n(\epsilon)=\min\{n \mid v_{n+1} < \epsilon\}$. 
Surely, @Adayah's proof of this lemma is the more straightforward one. As a comment too long for a comment, I just want to explain how to derive this lemma from a general result about integrability (finite expectation) of random variables. We can derive it from the two following facts: 
1) For any positive random variable $X$, 
$$E(X) = \int_0^1 F^{-1}(u)\mathrm{d}u,$$ 
where $F$ is the cumulative distribution function of $X$ and $F^{-1}$ is its left-continuous inverse (the inverse when it exists). 
Consequently, $$E(X) < \infty \implies \epsilon F^{-1}(1- \epsilon) \to_{\epsilon\to 0^+} 0.$$
And consequently, for any positive and right-continuous increasing function $H$, 
$$E(H(X)) < \infty \implies \epsilon H(F^{-1}(1- \epsilon)) \to_{\epsilon\to 0^+} 0.$$
(to prove this consequence, consider the right-continuous inverse of $H$ to check that $H\circ F^{-1}$ is the left-continous inverse of $H(X)$).
2) When $X$ is a discrete random variable, $$\sum_{n \geq 0} u_n\Pr(X>n) = E(\sum_{k=0}^X u_k).$$  (continuous version: $E(U(X))=\int u(x) \Pr(X>x)\mathrm{d}x$)
Then we can prove the lemma by firstly assuming without loss of generality that $v_0\leq 1$ and taking $X$ such that $\Pr(X>n)=v_n$. Then fact 2 gives $\sum u_n v_n = E(H(X))$ with $H(x) = \sum_{k=0}^x u_k$. Then it suffices to check that $n(\epsilon)=F^{-1}(1-\epsilon)$ and apply fact 1.
