Assume $\alpha \in (0,1), \beta > 0$. Consider this sum
\begin{align} s(n) = \sum_{i=0}^{n-1} \frac{\alpha^i}{(n-i)^\beta} \end{align}
Can we show an upper bound $b(n)$ on $s(n)$ that is decreasing in $n$ for some finite $\beta$ (e.g. $\beta = 2$)? Specifically, I want $\lim_{n \rightarrow \infty} b(n) = 0$.
Attempt: It is easy to see that the above sum is bounded independent of $\alpha$. For example for $\beta = 2$: $$ s(n) \leq \sum_{i=0}^{n-1} \frac{1}{(n-i)^2} \leq \sum_{i=1}^\infty \frac{1}{i^2} < 2. $$
Intuitively, it seems that $s(n)$ includes the sum of a few $1/n^\beta$ terms and the rest of the terms are small due to exponential decay $\alpha^i$.
Here is the plot for $\beta =1$:
Anyways, here is a simple proof for an upper bound.
Noticing the $(\alpha^i , 0 \le i \le n-1)$ is increasing in $i$ and $( \frac{1}{(n-i)^{\beta}}, 0 \le i \le n-1)$ is decreasing in $i$, by the Chebysev's inequality for monotonic sequences, we imply that: $$s(n) \le \frac{1}{n}\left( \sum_{i=0}^{n-1} \alpha^i \right)\left( \sum_{i=0}^{n-1} \frac{1}{(n-i)^{\beta}}\right) $$ So $$\begin{align} s(n) &\le \frac{1}{n}\frac{1}{1-\alpha}(1-\beta)n^{1-\beta}=\frac{1-\beta}{1-\alpha}n^{-\beta} \quad & \text{ if } \beta < 1 \\ s(n) &\le \frac{1}{n}\frac{1}{1-\alpha}\ln(n) &\text{ if } \beta = 1 \\ s(n) &\le \frac{1}{n}\frac{\beta}{1-\alpha} & \text{if } \beta >1 \end{align}$$