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what would be $\{s_n\ge0\}$ such that $$\sum_{n=1}^\infty s_n$$ converges

but

$$\lim_{n\to\infty}(n s_n) \neq 0$$

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    $\begingroup$ First, \ne will give you "not equal." You should try a sequence that does not decrease to $0$. $\endgroup$ – Ted Shifrin May 2 '13 at 17:54
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    $\begingroup$ @Ultra3: Please don't change the question like that (after getting multiple answers). Ask a new one. I am rolling it back. $\endgroup$ – Aryabhata May 3 '13 at 1:53
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$a_n=\frac{1}{n}$ if $n$ is a perfect square and $a_n=\frac{1}{n^2}$ otherwise.

P.S. This example is generic in the following sense:

Lemma If $a_n \geq 0, \sum a_n$ is convergent and $\lim_n na_n$ exists then $\lim na_n=0$.

Proof: Assume by contradiction that $\lim_n na_n \neq 0$. Then there exists some $M>0$ so that $na_n >M$ for all $n>N$. Then

$$a_n >M \frac{1}{n} \forall n >N \Rightarrow \sum_a_n =\infty$$ since the harmonic series diverges.


Thus, the only way you can construct an example as you want is by trying to make $\lim_n na_n$ not to exists.

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  • $\begingroup$ You beat me to it! It is instructive to see that examples exist in which all terms are strictly positive. $\endgroup$ – TonyK May 2 '13 at 18:05
  • $\begingroup$ @TonyK clark included a similar example, and I think he beat me to it, but I didn't see it before posting. His example is completely lost in his post, only saw it at the second reading.... $\endgroup$ – N. S. May 2 '13 at 18:09
  • $\begingroup$ One should state the problem differently. By $\lim_{n\to\infty}(n a_n) \neq 0$ I would mean: the limit exists but it is not zero. However the proposed problem needs: $na_n$ does not converge to zero. A different thing. $\endgroup$ – GEdgar May 2 '13 at 18:19
  • $\begingroup$ I don't understand how $\lim_{n\to\infty}(n a_n) $ is not 0. Because $n* \frac{1}{n^2} = \frac{1}{n}$ and the limit of that goes to 0. $\endgroup$ – Ultra3 May 2 '13 at 19:11
  • $\begingroup$ @Ultra3 But when $n$ is a perfect square, $n *a_n=n \frac{1}{n}=1 $ and it doesn't go to $0$. Formally let $k_n=n^2$. Then $a_{k_n}=\frac{1}{k_n}$ and for this subsequence $\lim_n (k_n a_{k_n})=1$. $\endgroup$ – N. S. May 2 '13 at 19:35
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An counter-example inlovning positive and negative terms is when

$a_n \frac{(-1)^n}{\sqrt{n}}$ the convegence can be check using the Alternating series test.

For an example with positive terms choose

$a_n= \frac{1}{n}$ when $n$ is perfect square otherwise let it be $0$

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  • $\begingroup$ +1, I think your answer was first, you should make the actual (positive) answer more visible, since I and most people probably miss it :) $\endgroup$ – N. S. May 2 '13 at 18:16
  • $\begingroup$ Yes, I did too! But it's not that important, it it? $\endgroup$ – TonyK May 2 '13 at 19:15
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Answers are up, this is more a meta-answer.

The problem is false as presented. The notation "$\displaystyle \lim_{n \to \infty} na_n \neq 0$" means that the limit exists and is nonzero, and if (as stated in the question) the terms of the sequence are positive, that implies $\sum a_n$ is infinite. The correct statement is "$na_n$ does not converge to zero", which includes the case where no limit exists.

The similarity between the posted answers is explained by the fact that in any solution there exists an infinite subsequence where $na_n$ is at least equal to a constant but for which $\sum a_n$ restricted to the subsequence is finite. For construction of examples, it is necessary and sufficient to specify such a subseries, and fill the rest of the terms with any convergent series.

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  • $\begingroup$ As far as I know that is not really a standard notation, even if it is the natural one. I saw textbooks using it both ways. While $\lim_{n \to \infty} na_n \neq 0$ seems to implicitly assume that the limit exists, one can also easily argue that $\lim_{n \to \infty} na_n \neq 0$ is the negation of $$\lim_{n \to \infty} na_n \eq 0$... But I agree, because of this issue the OP should clarify what he means by $\lim_{n \to \infty} na_n \neq 0$. $\endgroup$ – N. S. May 2 '13 at 20:06
  • $\begingroup$ I would fault the textbook in cases where the correctness of the problem depends on making the correct guess as to what the notation means, and there is no obvious reason to disfavor the wrong guess except discovering that it's wrong after trying to solve the problem. $\endgroup$ – zyx May 2 '13 at 20:16
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Let $a_n=(-1)^{n+1} \frac {1}{n},\, n=1, 2, 3, \ldots.$ Then $ \sum_{n=1}^{\infty} a_n =\ln 2$.

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  • $\begingroup$ $a_n$ is not positive. $\endgroup$ – N. S. May 2 '13 at 18:17
  • $\begingroup$ This does not address the question. $\endgroup$ – Pedro Tamaroff May 2 '13 at 20:54

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