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I thought because $p<1$ it would be divergent, but apparently not. Why is that?

user1551
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emk
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  • Hmm, what is $p$? – user1551 Jul 22 '13 at 23:23
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    what is $p$? Maybe you mean $\sum n^{-1/3}$? – sigmatau Jul 22 '13 at 23:23
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    Are you talking about the sequence $\dfrac{1}{n^{1/3}}$ which converges to $0$ or the series $\sum\dfrac{1}{n^{1/3}}$ which diverges? – Adriano Jul 22 '13 at 23:24
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    Why has this got so many downvotes? It looks like a perfectly valid question, albeit one arising from a misunderstanding. – Billy Jul 22 '13 at 23:25
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    @Billy we still don't know if it is a series or a sequence. The $p$ is undefined. There is no introduction and people are answering supposing it were a series without knowing it. – sigmatau Jul 22 '13 at 23:31
  • @AmireBendjeddou So? Help the OP to fix their question instead of just bashing the downvote button. Are you here to help, or not? – Billy Jul 23 '13 at 01:28
  • @Billy Yes i'm here to help , that's why i asked the OP to be clear (still no response).I usually never downvote a question , but if i do it's because i think i might be able to help but it's just not clear what the question really is. But your are right , we are here to help not to judge! – sigmatau Jul 23 '13 at 01:39
  • @AmireBendjeddou Ah, sorry, I didn't see that you were one of the above (helpful) commenters. But yes, I think that comments of your sort are good (the OP might not have responded because only 2 hours have passed!), and downvotes are kind of unproductive. I tend to find they drive new users away feeling attacked, rather than encouraging them to ask better questions. – Billy Jul 23 '13 at 01:44

5 Answers5

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Note that $n^{1/3}\to\infty$ as $n\to\infty$. And your sequence is $$n^{-1/3}=\frac{1}{n^{1/3}}$$

ADD It seems you're confusing things. For any $p>0$, $$n^{-p}\to 0$$

However $$\sum_{n\geqslant 1}n^{-p}$$ converges only when $p>1$.

Pedro
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The sequence $\{n^{-1/3}\}_{n=1}^\infty$ is convergent, since $n^{-1/3} = \frac{1}{n^{1/3}}$ tends to $0$ as $n$ grows. The series (i.e. sequence of partial sums) resulting in

$$\sum\limits_{n = 1}^\infty n^{-1/3}$$

is divergent, however.

2

You, certainly, know the series $\sum (1/n)$ and know that it is divergent. There is a nice approach in which we can test the divergence or convergence. That is the Quotient Test or Limit comparison test. According to it, if $$\lim_{n\to\infty}\frac{u_n}{v_n}=A\neq0, ~~\text{or}~~ A=\infty$$ then $\sum u_n$ and $\sum v_n$ have the same destiny. Here, take $\sum v_n=\sum (1/n)$.

Mikasa
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By the inequality $$\frac{1}{n}\leq \frac{1}{n^{1/3}}$$ we conclude that the serie $\displaystyle\sum_n \frac{1}{n^{1/3}}$ is divergent .

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The series $\displaystyle \sum_{n\geqslant 1} n^{-p}$ converges for $p>1$, if you are curious to why, you could search up Cauchy Condensation Test or even the Integral test.

Pedro
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WhizKid
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