# Prove $\lim\limits_{n\to\infty}na_n\ln(n)=0$

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.

-
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 $b_{n}$? –  user2468 Apr 21 '12 at 0:09
@J.D. I confess having no idea about the way your comment could help the OP. –  Did Apr 22 '12 at 11:48
@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! –  user2468 Apr 22 '12 at 15:12
@J.D. Nothing ensures that the sequence $(b_n)$ is nonincreasing (in case this is what you had in mind). –  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, $na_n\log n\to0$.

-
Thank you for answering some many questions of mine. Maybe I need to think for a rather long time before I post the question. –  89085731 Apr 22 '12 at 10:51

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

-
Your second application of the Cauchy condensation test is illicit since one does not assume that $(a_n)$ is monotone. –  Did Apr 22 '12 at 10:37
@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. –  Ragib Zaman Apr 22 '12 at 10:47
You are right. Sorry about the trouble. –  Did Apr 22 '12 at 10:51