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

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

• Use the Taylor expansion of $(1+x)^{1-a}$ with $x=1/(k+1)$ equivalently $\frac{1}{k^{a-1}} - \frac{1}{(k+1)^{a-1}} = \int_k^{k+1} (a-1)t^{-a}dt =(a-1)k^{-a}+(a-1) a\int_k^{k+1} \int_k^t (k^{-a-1}-u^{-a-1})dudt$ – reuns Apr 2 at 20:46