0
$\begingroup$

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.

  • 1
    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 $k \ge 0$.2011-10-18
  • 1
    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}}$2011-10-18
  • 0
    @Ross Millikan, I mean the sequence. I know how to solve if it is the series...2011-10-18
  • 0
    @aengle, could you explain in detail since I'm just a beginner in analysis. Thank you for your help~2011-10-18
  • 0
    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.2011-10-18
  • 0
    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

1 Answers 1

4

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?