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Let $\{a_n\}$ be a sequence such that

  • $a_n\geq 0$ for all $n$
  • $\{a_n\}$ is monotonically decreasing
  • $\sum_{n=1}^\infty a_n$ converges

Is it true that as $n\rightarrow\infty$ then $$n\log n\;a_n\rightarrow 0$$

Given the hypotheses, we can show that $n a_n\rightarrow 0$ as $n\rightarrow\infty$. This follows since $$0\leq 2na_{2n}\leq 2(S_{2n}-S_n)\quad\text{ and }\quad 0\leq(2n+1)a_{2n+1}\leq 2(S_{2n+1}-S_n)+a_{2n+1}$$

I've been trying to adapt this approach to $n\log n\; a_n$, but it has been fruitless so far. I've been working with inequalities that involve $\log$, but they each seem to be too 'weak'; in that, I end up with a product sequence with one part going to $0$ and the other going to $\infty$. I also tried condensation, but I can't determine whether the general term of that new series forms a monotonically decreasing sequence. I also have not been able to come up with a counter-example.

Any help in resolving the question in either direction is appreciated.

UPDATE

RRL's approach solves every case where $\lim\inf n\log n\;a_n>0$, but we still haven't resolved the case where $\lim\inf n\log n\; a_n=0$ and $\lim\sup n\log n\; a_n>0$.

On a serendipitous note, I was reading through one of my books on analysis and the result for $na_n\rightarrow 0$ was posed as a problem. It came with a footnote that $1/n$ cannot be replaced by a function that approaches $0$ quicker. I take this to include $1/n\log n$. However, I'm having trouble finding the exact reference the author cites. The book I found this in is "Elementary Real and Complex Analysis" by Geogi E. Shilov (first printed in 1973), and the only direction he gives is the name "A.S. Nemirovski". So, any help directing me to this reference would be a big help as well.

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Start with the case where the limit exists, but $n \log n \,a_n \to L > 0$. Then $a_n > L/ (2n \log n)$ for sufficiently large $n$, and $\sum a_n$ must diverge.

Hence, if the limit exists it must equal 0.

As observed by Bryan, if the limit does not exist, but $\liminf (n \log n \, a_n) = l > 0$, then again we have $a_n > l/ (2n \log n)$ for sufficiently large $n$, and $\sum a_n$ must diverge.

There remains the case where $\liminf (n \log n \, a_n) = 0$ and $\limsup (n \log n \, a_n) > 0$. For the conjecture to be true, we must show that this is impossible when $\sum a_n$ is convergent. Any hope of proving this rests on the assumption that $a_n$ is decreasing.

If $a_n$ is not monotone then we have a counterexample where $a_n = 1/(n \log n)$ for $n = j^2$ and, otherwise, $a_n = 1/n^2$. Here we have $\sum a_n$ convergent, but

$$0 = \liminf (n \log n \, a_n) < \limsup (n \log n \, a_n) = 1.$$

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  • $\begingroup$ Thank you. If the limit is $+\infty$ same problem. If the limit doesn't exist consider unbounded and bounded image separately. Take a subsequence going to infinity in the former and use Bolzano-Weierstrass in the latter. This approach shows that $n\log n\log\log n\; a_n\rightarrow 0$ and so forth as well which was going to be my next attempt. $\endgroup$ – Robert Wolfe Mar 22 '15 at 20:30
  • $\begingroup$ Actually using subsequences lead to other difficulties. We don't know that the series built out of this subsequence $\sum 1/n_k\log n_k$ will diverge still. If we knew that $\lim\inf n\log n a_n>0$, we'd be fine. But there's still a problem when $\lim\inf =0$ and $\lim\sup >0$. $\endgroup$ – Robert Wolfe Mar 22 '15 at 21:00
  • $\begingroup$ @Bryan - I expanded the answer to show that the final case must make use of the monotonicity condition. $\endgroup$ – RRL Mar 24 '15 at 2:13
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This morning I looked up some references regarding the issue being discussed here. I don't have time to read over and discuss the results, but maybe the references will be of use to you and anyone else interested.

See Article 9 on pp. 26-27 of Bromwich [1] (= Article 9 on pp. 30-31 of the 1925 2nd edition) and Article 19 on pp. 25-26 of Hobson [2]. For historical information about this result, see around p. 228 of Molk/Pringsheim [3] (originally written in German by Pringsheim, and later translated to French and slightly extended by Molk). From what I can tell, Pringsheim [4] is the main origin of the results discussed in the references I just cited, but it's likely that Russian and other authors may have played a role, some possibly known to the above authors and some possibly not known to the above authors.

[1] Bromwich, An Introduction to the Theory of Infinite Series, 1908.

[2] Hobson, The Theory of Functions of a Real Variable and the Theory of Fourier's Series, Volume II, 2nd edition (latest), 1926.

[3] Molk/Pringsheim, Chapter 4 in Encyclopédie des Sciences Mathématiques Pures et Appliquées, 1904. alternate internet location 1 and alternate internet location 2 and alternate internet location 3

[4] Pringsheim, Allgemeine Theorie der Divergenz und Convergenz von Reihen mit positiven Gliedern, Mathematische Annalen 35 (1890), 297-394. alternate internet location

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  • $\begingroup$ Very helpful. I looked at Bromwich (believe it or not I have the 1908 first edition). The note states that Pringsheim gives examples where $\lim a_nD_n = 0$ is not a necessary condition (for decreasing $a_n$ and $D_n$ growing more rapidly than $n$). Then I discovered this answer by Andre Nicolas: math.stackexchange.com/a/1205542/148510 where he constructs such sequences $(a_n)$ and $(D_n)$. However, this requires $(D_n)$ to be a sequence with blocks of very rapidly increasing integers of rapidly increasing length. Don't think we can show this for $D_n = n \log n$. $\endgroup$ – RRL Mar 27 '15 at 6:23

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