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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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    Thank you very much, @André Nicolas! Actually, as you pointed out, there was a typo... Sorry for that. I am asking about the sequence, not the series.2011-10-18

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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?