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As far as I can think of, the only conditionally convergent series are where $\sum na_n$ diverges. eg. $$\frac{(-1)^n}{n}$$ is conditionally convergent and $na_n = (-1)^n$ is divergent.

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Hint: $$\sum_{n = 2}^{\infty} \frac{1}{n \log n}$$ is divergent by the integral test.

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  • $\begingroup$ im unsure how this would help as ∑ n=2 to infinity seems conditionally convergent but it is divergent and we need ∑nan to conditionally converge $\endgroup$ – Joemans Mar 17 '16 at 1:21
  • $\begingroup$ No, the series I wrote is divergent. You'll need to make a suitable modification that's suggested by the series you wrote in your question to make it work. $\endgroup$ – user296602 Mar 17 '16 at 1:23
  • $\begingroup$ Honestly thank you so much, but that didnt help me at all. $\endgroup$ – Joemans Mar 17 '16 at 1:38
  • $\begingroup$ The example you gave is conditionally convergent because it's alternating. My series is not alternating. Change that fact by modifying my series. $\endgroup$ – user296602 Mar 17 '16 at 3:03

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