Show $\displaystyle \left(\frac{1}{n^k}\right)_{n\in\mathbb{N}$ is convergent iff. $\displaystyle k\geq 0$, and that $\displaystyle \left(\frac{1}{n^k}\right)\rightarrow 0, \ \forall k>0$

(i) assume $\displaystyle \left(\frac{1}{n^k}\right)$ is convergent.

Because it is convergent, there is some value alpha such that:

$\displaystyle \left(\frac{1}{n^k}\right)\rightarrow\alpha$

$\displaystyle \Rightarrow\left |\frac{1}{n^k}-\alpha\right |<\epsilon$

I am not sure if I am on the right track, and if I am, what should I do next?