Prove that $\sum{\frac{1}{(k+1)^\alpha}\frac{1}{(n+1-k)^\beta}}\le K\frac{1}{(n+1)^\alpha}$ 
I want to prove that
$$\sum_{k=0}^{n}{\frac{1}{(k+1)^\alpha}\frac{1}{(n+1-k)^\beta}}\le K\frac{1}{(n+1)^\alpha}$$
for all $n\ge 0$, where $1<\alpha\le\beta$ and $K$ a constant.

Everything I tried to do, I always arrived at the constant depending on $n$, which cannot happen. And I didn't want to use induction, is it possible?

For all $n\geq 1$ you have
$$\sum_{k=0}^n \frac{(n+1)^\alpha}{(k+1)^\alpha(n+1-k)^\beta} \leq \sum_{k=0}^n \frac{(n+1)^\alpha}{(k+1)^\alpha(n+1-k)^\alpha} = \sum_{k=0}^n \frac{1}{\left[(k+1)\left(1-\frac{k}{n+1}\right)\right]^\alpha} \, .$$
Since the summand is symmetric about $k=n/2$, where it acquires a unique minimum, it suffices to bound
$$\sum_{1 \leq k \leq n/2} \frac{1}{\left[(k+1)\left(1-\frac{k}{n+1}\right)\right]^\alpha} \leq \int_0^{n/2} \frac{{\rm d}k}{\left[(k+1)\left(1-\frac{k}{n+1}\right)\right]^\alpha} \\
\leq \frac{1}{\left(1-\frac{n}{2(n+1)}\right)^\alpha} \int_0^{\infty} {\rm d}k \, (k+1)^{-\alpha} \stackrel{\alpha>1}{\leq} \frac{2^{\alpha}}{(\alpha-1)} \, ,$$
since $n/(n+1)$ is increasing in $n$.
