Show that the series $\sum_{n=3}^{\infty} n(\log n)(\log {\log n})^2 a_n$ diverges. Suppose that the series $\sum_{n=1}^{\infty} a_n$ converges conditionally. Show that the series $\sum_{n=3}^{\infty} n(\log n)(\log {\log n})^2 a_n$ diverges.
Any hint on how to proceed?
 A: Suppose that $\sum_{n=3}^{\infty} n(\log n)(\log {\log n})^2 a_n$ converges. Then we have 
$$n(\log n)(\log {\log n})^2 a_n\to 0,$$
so 
$$|a_n|<\frac{1}{n(\log n)(\log {\log n})^2}$$
for large $n$. But by Cauchy condensation test, the series 
$$\sum_{n=3}^\infty \frac{1}{n(\log n)(\log {\log n})^2}<\infty,$$
which implies that $\sum a_n$ is absolutely convergent, contradiction.
A: Suppose $\sum_{n=3}^{\infty} n \log(n) (\log(\log(n)))^2 a_n < \infty$. Then $b_n := n \log(n) (\log(\log(n)))^2 a_n$ has $b_n \to 0$. This means that for all large enough $n$, we have $|b_n| < 1$, hence $$|a_n| < \frac{1}{n \log(n) (\log(\log(n)))^2}$$ Now, note that the function $f(x) = 1/(x \log(x) (\log(\log(n)))^2$ has the antiderivative $-1/\log(\log(x))$, and in particular, $\int_{3}^{\infty} f(x)$ converges. Since $f(x)$ is a monotone decreasing function for large $x$, the integral test implies $\sum_{n=3}^{\infty} f(n) $ converges. But then $\sum_{n=3}^{\infty} |a_n|$ converges by comparison, so $\{a_n\}$ is absolutely convergent, contradiction.
