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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?

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

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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.

Proof. (modeled after guest's answer)

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 $.

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