$\lim_{n\rightarrow\infty}n(\ln n)a_n=0$ implies $\sum a_n$ converges? Is it true that
If $$\lim_{n\rightarrow\infty}n(\ln n)a_n=0,$$
then the series
$$\sum_{n=1}^\infty a_n$$
converges?
If so, I want to know the proof.
If not, I want to know the counter example.
 A: The series $$\sum_{n=4}^\infty \frac{1}{n\log n \log\log n}$$ does not converge.  Use the integral test.
A: Assume that $\sum_{n=1}^\infty b_n$ is a divergent positive series. Then you can always find another divergent positive series $\sum_{n=1}^\infty a_n$ with the property that
$$
\lim_{n\to\infty} \frac{a_n}{b_n} = 0.
$$
(In your case, $b_n = \dfrac{1}{n\log n}$.)
One way to see this is to use a theorem of Abel and Dini: 

Theorem Assume that $\sum_{n=1}^\infty b_n$ is a divergent positive series,
  and let $B_n = b_1 + \cdots + b_n$ denote its partial sums. Then the
  series $$ \sum_{n=1}^\infty \frac{b_n}{B_n} $$ also diverges.

Proof First note that
$$
\frac{b_{n+1}}{B_{n+1}} + \frac{b_{n+2}}{B_{n+2}} +
 + \cdots + \frac{b_{n+k}}{B_{n+k}} \ge
\frac{b_{n+1}+b_{n+2}+\cdots+b_{n+k}}{B_{n+k}} = 1 - \frac{B_n}{B_{n+k}}.
$$
Since we assume that $B_n \to \infty$, for each $n$, we can choose $k_n$ such that
$$
\frac{B_n}{B_{n+k_n}} < \frac12,
$$
i.e.
$$ 
\frac{b_{n+1}}{B_{n+1}} + \frac{b_{n+2}}{B_{n+2}} +
 + \cdots + \frac{b_{n+k_n}}{B_{n+k_n}} > \frac12.
$$
Summing blocks like these, we see that
$$
\sum_{n=1}^\infty \frac{b_n}{B_n}
$$
diverges.
