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Suppose that the series $\sum_{n=1}^{\infty} a_n$ converges conditionally. Show that the series $\sum_{n=3}^{\infty} n(\log n)(\log {\log n})^2 a_n$ diverges.

Any hint on how to proceed?

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closed as off-topic by Namaste, Jendrik Stelzner, Delta-u, rtybase, A. Pongrácz Sep 3 '18 at 22:23

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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.

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  • $\begingroup$ This is exactly how I solved this problem when it was on Indiana's qual last January. +1 $\endgroup$ – Alfred Yerger Aug 14 '18 at 0:47
  • $\begingroup$ You beat me to it. +1 $\endgroup$ – Clement C. Aug 14 '18 at 0:50
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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.

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