# Show that the series diverges.

Suppose that the series $$\sum_{n=1}^{\infty} a_{n}$$ converges conditionally. Show that the series $$\sum_{n=3}^{\infty}n(\log n)(\log\log n)^{2} a_{n}$$ is divergent. I tried this question with summation by parts but no conclusion. Can anyone give some hint?

• What would $n(\log n)(\log \log n)^2a_n \to 0$ imply for $\sum a_n$? – Daniel Fischer Dec 10 '19 at 14:51
• Do you mean by contradiction? – user786 Dec 10 '19 at 14:53

Let series $$\sum n(\log n)(\log(\log n))^2a_n$$ be convergent. Then, we have $$\lim_{n\to\infty} n(\log n)(\log(\log n))^2a_n=0$$ implying $$\lim_{n\to\infty} n(\log n)(\log(\log n))^2|a_n|=\lim_{n\to\infty}\frac{|a_n|}{\frac{1}{n(\log n)(\log(\log n))^2}}=0.$$ This implies that series $$\sum |a_n|$$ is convergent by limit comparison test and by the convergence of the series $$\sum \frac{1}{n(\log n)(\log (\log n))^2},$$ which contradicts with conditional convergence of $$\sum a_n$$. Hence, series $$\sum n(\log n)(\log(\log n))^2a_n$$ is divergent.
If $$\sum a_n$$ converges conditionally, then $$\sum |a_n|$$ diverges.
Let $$b_n$$ be any positive sequence such that $$\sum \dfrac1{b_n}$$ converges.
Claim: $$\sum b_n a_n$$ diverges.
If $$\sum b_n a_n$$ converges, then $$b_n a_n \to 0$$, so that $$b_n |a_n| \to 0$$, or $$\dfrac{|a_n|}{\dfrac1{b_n}} \to 0$$. Therefore, since $$\sum \dfrac1{b_n}$$ converges, $$\sum |a_n|$$ also converges, which contradicts the conditional convergence of $$\sum a_n$$.