real analysis - Bounding the Series $sum_{k=1}^infty k^a frac{1}{k(k+1)}$

Let $0 < a < 1$. I'm trying to figure out whether the following series converges:

$$\sum_{k=1}^\infty k^a \frac{1}{k(k+1)}.$$

Now it's clear that if $a$ were greater than or equal to $1$ then this series would diverge since

$$\sum_{k=1}^\infty k^1 \frac{1}{k(k+1)} = \sum_{k=1}^\infty \frac{1}{(k+1)} = \sum_{k=2}^\infty \frac{1}{k} = \infty.$$

So this makes it a bit hard to think of a bound for the series in question. Any advice?

Answer

$$0\leq \frac{k^\alpha}{k(k+1)}\leq \frac 1{k^{2-\alpha}},$$
and $2-\alpha>1$.