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It is an elementary mathematical analysis problem, but I have some problem solving that.

Show that the sequence $1/n^k$ ,where $n \in \mathbb{N}$ is a natural number, is convergent if and only if $ k \geq 0$, and that the limit is $0$ for all $k > 0$.

I really don't know how to prove that when $ 0\leq k < 1$, the sequence is convergent.

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Do you want the sequence $1/n^k$ or the series $\sum 1/n^k$? In your first you say it (whichever it is) converges only for $k \ge 1$ and in the last it is convergent for any $k \ge 0$. – Ross Millikan Oct 18 '11 at 3:06
We know that $\dfrac{1}{n^k}, k\geq 0$ is monotone and bounded. We can respond explicitly to the $\epsilon$ challenge by setting $N = \dfrac{1}{\epsilon^{1/k}}$ – mathmath8128 Oct 18 '11 at 3:17
@Ross Millikan, I mean the sequence. I know how to solve if it is the series... – Winston Oct 18 '11 at 3:27
@aengle, could you explain in detail since I'm just a beginner in analysis. Thank you for your help~ – Winston Oct 18 '11 at 3:30
There may be a typo in your question. And there was one in this comment, that was pointed out by @Srivatsan Narayanan! The sequence $(1/n^k)$ converges if and only if $k\ge 0$. Apart from the trivial case $k=0$, the sequence converges to $0$. The series $\sum \frac{1}{n^k}$ converges if and only if $k>1$. Note it is not $k \ge 1$. My feeling is that in fact you are not asking about the series at all. Can you clarify? Someone will supply a full answer shortly after that. – André Nicolas Oct 18 '11 at 3:31

If $k=0$, then $1/n^k$ is constant $1$, so convergent. If $k>0$, it converges to $0$ as you say. To prove this, for a given $k$ which is greater than $0$, if I give you an $\epsilon >0$ you need to find an $N$ so that for all $n>N,\ 1/n^k<\epsilon$. As $1/n^k$ always decreases with $n$, all you have to do is find an $N$ where it is certainly below. Can you do that?

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