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Let $a_n>0$,$\sum a_n$ is convergent,$na_n$ is monotone, Prove: $$\lim\limits_{n\to\infty}na_n\ln(n)=0$$

I try to prove $a_n=o(n\ln(n))$, but it doesn't work.


marked as duplicate by davidlowryduda Sep 28 '17 at 18:54

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  • $\begingroup$ Let $b_n = n a_n \ln n,$ and $b_{n+1} = (n+1) a_{n+1} \ln (n+1).$ What is the relationship between $b_{n+1}$ and $b_{n}$? $\endgroup$ – user2468 Apr 21 '12 at 0:09
  • $\begingroup$ @J.D. I confess having no idea about the way your comment could help the OP. $\endgroup$ – Did Apr 22 '12 at 11:48
  • $\begingroup$ @Didier I guess I was thinking out loud about how the sequence is decreasing.. but you're right; it is not really a helpful comment! $\endgroup$ – user2468 Apr 22 '12 at 15:12
  • $\begingroup$ @J.D. Nothing ensures that the sequence $(b_n)$ is nonincreasing (in case this is what you had in mind). $\endgroup$ – Did Apr 22 '12 at 15:19

Since $(na_n)$ is positive and monotone, $(na_n)$ is nonincreasing and converges to zero hence $na_n=\sum\limits_{k\geqslant n}c_k$ for a nonnegative summable sequence $(c_k)$.

Since $\sum\limits_na_n$ converges, $\sum\limits_kc_kx_k$ converges, with $x_k=\log k$. Furthermore, $na_n\log n=\sum\limits_{k}c_kx_k(n)$ with $x_k(n)=[n\leqslant k]\log n$. Hence $x_k(n)\leqslant x_k$ and $x_k(n)\to0$ when $n\to\infty$. By dominated convergence, $na_n\log n\to0$.

  • $\begingroup$ Thank you for answering some many questions of mine. Maybe I need to think for a rather long time before I post the question. $\endgroup$ – 89085731 Apr 22 '12 at 10:51
  • $\begingroup$ I'm stuck as to why $\sum a_k$ converges implies $\sum c_kx_k$ converges. Would you care to elaborate a bit? $\endgroup$ – Vim Jun 26 '15 at 14:19
  • 1
    $\begingroup$ @Vim $$\sum_na_n=\sum_n\frac1n\sum_{k\geqslant n}c_k=\sum_kc_k\sum_{n\leqslant k}\frac1n\sim\ldots$$ $\endgroup$ – Did Jun 26 '15 at 17:10
  • $\begingroup$ @Did thank u so much. $\endgroup$ – Vim Jun 26 '15 at 17:17

Lemma: If $\sum b_n $ converges and $b_n>0$ is monotone, then $n b_n \to 0 .$

Proof: By the Cauchy Condensation test we have $\sum 2^n b_{2^n} $ converges as well, so $2^n b_{2^n}\to 0.$ For $2^n < k < 2^{n+1} $ we have $k b_k \leq 2^{n+1} b_{2^n} \to 0.$ Thus $n b_n \to 0.$

By the Cauchy Condensation test $\sum 2^n a_{2^n} $ converges, the lemma shows $ 2^n a_{2^n} \to 0$ and we are given that this occurs monotonically.

Thus applying our lemma to $\sum 2^n a_{2^n} $ gives us that $ n 2^n a_{2^n} \to 0.$ Now for $2^n < k < 2^{n+1} $ we have $$ \log_2 k\cdot k a_k \leq (n+1) 2^{n+1} a_{2^n} \to 0.$$

Thus $$ \lim_{n\to\infty} n a_n \log n = 0.$$

  • $\begingroup$ Your second application of the Cauchy condensation test is illicit since one does not assume that $(a_n)$ is monotone. $\endgroup$ – Did Apr 22 '12 at 10:37
  • $\begingroup$ @Didier We are given $(na_n)$ is monotone, and if it were increasing then $\sum a_n$ wouldn't converge so it must be decreasing. So $ (n+1)a_{n+1} \leq n a_n. $ Dividing through by $n$ gives $ \frac{n+1}{n} a_{n+1} \leq a_n .$ Clearly $ a_{n+1} \leq \frac{n+1}{n} a_{n+1} $ so we have $ a_{n+1} \leq a_n $ as well. $\endgroup$ – Ragib Zaman Apr 22 '12 at 10:47
  • $\begingroup$ You are right. Sorry about the trouble. $\endgroup$ – Did Apr 22 '12 at 10:51

I think this one is also a nice answer:

Lets prove that $\lim\sup n a_n \log n = 0$. For this, we note that, by the basic estimates on the partial harmonic sums,

$$ \limsup n a_n \log n = \limsup n a_n \sum_{k=k_0}^n \frac{1}{k} $$

For all $k_0 \ge 0$. As $\sum a_n < + \infty$, we must have that, from the hypothesis, $na_n$ is decreasing to zero. But then $a_n/k \le a_k/n$ for $k\le n$. Using this on the $\limsup$ above, we see that

$$ \limsup n a_n \log n \le \limsup \sum_{k=k_0}^n a_k = \sum_{k=k_0}^{\infty} a_k $$

As $k_0$ was arbitrary, and as the sum converges, we conclude the desired result.


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