Bound for $K_n=\sum_{i=1}^{n-1}\dfrac{1}{i^{(\alpha+1)}}.\dfrac{1}{(n-i)^{(\alpha+1)}}, \alpha>1$

Question

I would like to bound the sum $$K_n=\sum_{i=1}^{n-1}\dfrac{1}{i^{(\alpha+1)}}.\dfrac{1}{(n-i)^{(\alpha+1)}}.$$ My attempt yield: $$K_n \leq \dfrac{1}{(n-1)^{\alpha}}+\dfrac{1}{(n-1)^{\alpha}} \leq \dfrac{1}{(n-1)^{\alpha}}.$$

However, running some numerical simulation on $\alpha = 1, 1.5, 1.7, 2: n^{\alpha + 1}K$, seems bounded, $n^{\alpha + 1-0.01}K$ is decreasing in n for n up to $10^7$, though decreases very slowly $n^{\alpha + 1 - 1/2}K$, decreases rapidly for n up to $10^7$

The numerics would suggest that $n^{-\alpha}$ is not optimal as a bound for $K$ and perhaps $n^{-(\alpha + 1)}$ is.

Answer 1

Let $$K_n=\sum_{i=1}^{n-1}\dfrac{1}{i^{(\alpha+1)}}.\dfrac{1}{(n-i)^{(\alpha+1)}}$$ We have $$K_n < \sum_{i=1}^{n-1} \left( \dfrac{1}{i^{(\alpha+1)}}+\dfrac{1}{(n-i)^{(\alpha+1)}}\right)\frac{1}{ (n/2)^{\alpha+1}} < \frac{C}{n^{\alpha+1}}$$ for $$C=2^{\alpha+1}\sum_{i = 1}^{\infty} \frac{2}{i^{\alpha+1}}$$