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$$\sum_{n =1}^{\infty}\sin\left(\frac{\pi\cdot n}{4}\right)\cdot \sqrt[9]{\ln\left(\frac{n+12}{n+9}\right)}$$ How to find convergence of this series? I researched the absolute convergence and get $$\exp^{\frac{1}{3\cdot (n+9)}}$$ Thanks a lot!

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    $\begingroup$ your series does converge $\endgroup$ – Dr. Sonnhard Graubner May 20 '17 at 14:56
  • $\begingroup$ Please state what have you tried... $\endgroup$ – Alex Vong May 20 '17 at 14:56
  • $\begingroup$ @AlexVong I researched the absolute convergence and get $\exp^{\frac{1}{3\cdot (n+9)}}$ $\endgroup$ – user448072 May 20 '17 at 15:06
  • $\begingroup$ @Dr.SonnhardGraubner Can you explain why does it converge? $\endgroup$ – user448072 May 20 '17 at 15:09
  • $\begingroup$ What do you mean by "I researched ..." and got that? Makes no sense. Surely you have some thoughts on this. Come on, what does the sequence $\sin(\pi n/4)$ look like? $\endgroup$ – zhw. May 20 '17 at 15:38
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Hints: The series does not converge absolutely. Proof idea: Sum over the indices $n=2,10,18,26,\dots.$ The series does converge conditionally. Proof idea: The partial sums of $\sum \sin(\pi n/4)$ are bounded and the terms $[\ln ((n+2)/(n+9))]^{1/9}$ decrease to $0.$

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  • $\begingroup$ How to proof that series doesn't converge absolutely? $\endgroup$ – user448072 May 20 '17 at 16:00
  • $\begingroup$ What happens when you sum over the indices I mentioned? $\endgroup$ – zhw. May 20 '17 at 16:16
  • $\begingroup$ I've already solved this example, thanks $\endgroup$ – user448072 May 20 '17 at 16:19

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