Estimate for discrete convolutions

Question

Let $\alpha,\beta\in \mathbb R$, and define the expression: $$ A_n:=\sum_{m=1}^n (n+1-m)^\alpha m^\beta. $$ What is the order of $A_n$? Do we have $A_n\sim n^{\alpha+\beta+1}$?

Answer 1

When we divide $A_{n}$ by $(n+1)^{\alpha+\beta+1}$, we have $$ \frac{A_{n}}{(n+1)^{\alpha+\beta+1}} = \frac{1}{n+1}\sum_{m=1}^{n}\left(1-\frac{m}{n+1}\right)^{\alpha}\left(\frac{m}{n+1}\right)^{\beta} $$ and RHS converges to $$ \int_{0}^{1}(1-x)^{\alpha}x^{\beta} dx = B(\alpha+1, \beta+1) $$ which is a Beta function. Hence we have $A_{n}\sim n^{\alpha+\beta+1}$.