If the sequence $\{{1\over n^k}\}$ where $n\in \mathbb{N}$ is convergent, then $k\geq 0$ and the limit $0$ for all $k>0$. If the sequence $\{{1\over n^k}\}$ where $n\in \mathbb{N}$ is convergent, then $k\geq 0$ and the limit $0$ for all $k>0$.
What I have:
Assume that $k<0$, need to show that this contradicts the fact that $\{{1\over n^k}\}$ converges. Any help is greatly appreciated, thank you.
 A: If $k<0$ then for every $M>0$ we can choose $N\in\mathbb{N}$ such that $N>M^{-1/k}$. Then, if $n>N$ then
$$n^{-k}>N^{-k}>M$$
This shows that $n^{-k}$ is divergent to $\infty$.
Now you need to show that when $k\geq0$ it converges to $0$. The procedure is similar, but trying to push down instead of up.
A: Suppose the sequence $\lbrace \dfrac{1}{n^{k}}\rbrace $ is convergent. To obtain a contradiction assume that $ k<0 $. Then there exists $ t>0 $ such that $ k=-t $. Then $ \dfrac{1}{n^{k}}=n^{t} $ for each $ n\in \mathbb{N} $. Since $\lbrace \dfrac{1}{n^{k}}\rbrace $ is convergent there exists $ x\in \mathbb{R} $ such that for each $ \epsilon >0 $, there exists $ n_{\epsilon}\in \mathbb{N} $ such that for each $ n>n_{\epsilon} $, $ \vert \dfrac{1}{n^{k}}-x\vert <\epsilon $.
choose $ \epsilon =1 $. Let $ N\in \mathbb{N} $. Now choose $ n\in \mathbb{N} $ so that $ n>N $ and $ n^{t}>x+1 $. Therefore there exists $ \epsilon _{0}(=1)>0 $ such that for each $ N\in \mathbb{N} $, there exists $ n>N $ such that $ \vert n^{t}-x\vert >\epsilon _{0} $. This is a contradiction. Therefore $ k\geq 0 $.
To show $\lbrace \dfrac{1}{n^{k}}\rbrace $ is convergent 0, Let $ \epsilon >0 $. Choose $ N\in \mathbb{N} $ such that $ \dfrac{1}{N^{k}}<\epsilon $. Then for each $ n>N $ we have that $ \vert \dfrac{1}{n^{k}}-0\vert = \dfrac{1}{n^{k}} <\epsilon $. Therefore $\lbrace \dfrac{1}{n^{k}}\rbrace $ is convergent 0. $ \square $ 
