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.
 A: 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 &lt k &lt 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 &lt k &lt 2^{n+1} $ we have $$ \log_2 k\cdot  k a_k \leq (n+1) 2^{n+1} a_{2^n} \to 0.$$
Thus $$ \lim_{n\to\infty} n a_n \log n = 0.$$
A: 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, $na_n\log n\to0$.
A: 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. 
