Suppose $\{r_i\}_{i\in \mathbb{N}} \subseteq (0,1)$ is a $p$-summable sequence for some $p\geq 1$ (namely $\sum_{i\in \mathbb{N}}r_i^p < \infty$). In general, depending on the choice of $p$ or not, can we find $\lambda\in (0,1)$ such that $\lambda^n r_1 < r_n$ whenever $n$ is large enough?
What inspired this question is the following: if $\lim_i \dfrac{r_{i+1}}{r_i} < 1$ the ratio test is passed, regardless of the choice of $p$, by putting $\lambda = \lim_i \dfrac{r_{i+1}}{r_i}$ it is clear that for any $n\in\mathbb{N}$ that is large enough, we have $r_n < \lambda r_{n+1}$, or the decreasing rate of $r_n$ is strictly less than $\lambda$. In this case we have $\dfrac{r_1}{r_n} > \lambda^{1-n}$, or the diverging rate of $\dfrac{r_1}{r_n}$ is greater than $\lambda^{-1}$, When the result of the ratio set is not clear, or $\lim_i \dfrac{r_{i+1}}{r_i} = 1$, I wonder if there is a $\lambda\in (0, 1)$ such that the diverging rate of $\dfrac{r_1}{r_n}$ is less than $\lambda^{-1}$. I believe the choice of $p$ will affect the decreasing rate of $r_n$ but do not know how.