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?

Complement to Diger's proof:
According to Diger's idea, we have
\begin{align*}
 &\sum_{k=0}^{n}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\beta}}\\
 \le\,& \sum_{k=0}^{n}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\alpha}}\\
 \le\,& \sum_{0\le k\le n/2}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\alpha}}
 + \sum_{n/2\le k\le n}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\alpha}}\\
 =\,& \sum_{0\le k\le n/2}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\alpha}}
 + \sum_{0\le m\le n/2}{\frac{1}{(n-m+1)^\alpha}\frac{(n + 1)^\alpha}{(m + 1)^\alpha}}\\
 =\,& 2\sum_{0\le k\le n/2}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-k)^\alpha}}\\
 \le\,& 2\sum_{0\le k\le n/2}{\frac{1}{(k+1)^\alpha}\frac{(n + 1)^\alpha}{(n+1-n/2)^\alpha}}\\
 \le\,& 2^{\alpha + 1}
 \sum_{k=0}^ {\infty}\frac{1}{(k+1)^\alpha}.
\end{align*}
Since $\sum_{k=0}^ {\infty}\frac{1}{(k+1)^\alpha}$ is convergent (p-series test), the desired result follows.
