show that $\frac{1}{k^{a-1}} - \frac{1}{(k+1)^{a-1}} \geq \frac{1}{\zeta(a).k^a}c $

Question

I want to Apply the acceptance reject method to the zipf distribution. For that i want to use q(k)= $\frac{1}{k^{a-1}} - \frac{1}{(k+1)^{a-1}}$

I have to show there exist c>1, such that $\frac{1}{k^{a-1}} - \frac{1}{(k+1)^{a-1}} \geq \frac{1}{\zeta(a).k^a}c $

where $\zeta(a)= \sum_{n\geq1} \frac{1}{n^a}$ , a>1