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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.

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    @J.D. Nothing ensures that the sequence $(b_n)$ is nonincreasing (in case this is what you had in mind).2012-04-22

3 Answers 3

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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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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.$

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    You are right. Sorry about the trouble.2012-04-22
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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, $na_n\log n\to0$.

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    @Did thank u so much.2015-06-26