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This series is not absolutely convergent because \begin{align*} \lim_{k\rightarrow+\infty}\frac{\bigl|\left(-1\right)^{k+1}\sin\left(\frac{1}{k}\right)\bigr|}{\frac{1}{k}} & =\lim_{k\rightarrow+\infty}\frac{\bigl|\sin\left(\frac{1}{k}\right)\bigr|}{\frac{1}{k}}\\ & =\lim_{k\rightarrow+\infty}\frac{\sin\left(\frac{1}{k}\right)}{\frac{1}{k}}\\ & =1 \end{align*} Since $ \sum_{k=1}^{\infty}\frac{1}{k} $ is divergent, so is not absolutely convergent.

Is this conditionally convergent?

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Yes, it converges conditionally, by the Leibniz criterion.

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  • $\begingroup$ Leibniz's criterion $\endgroup$ – user574380 Jul 13 '18 at 23:10
  • $\begingroup$ Is this correct? Because the derivative is $ \left(-1\right)\frac{1}{x^{2}}\cos\left(\frac{1}{x}\right) $, so when x from 1 and goes to positive infinity, the cos term is always positive. So $ \sin\left(\frac{1}{k}\right) $ is monotone decreasing. $\endgroup$ – Lithium Jul 13 '18 at 23:10
  • $\begingroup$ Sure, $\left(\sin\left(\frac1k\right)\right)_{k\in\mathbb N}$ is decreasing. Is that a problem? It's one of the hypothesis of the Leibniz criterion. $\endgroup$ – José Carlos Santos Jul 13 '18 at 23:13
  • $\begingroup$ May I ask you that why this one is conditionally convergent? $ \sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{k^{k}}{\left(k+1\right)^{k+1}} $ $\endgroup$ – Lithium Jul 13 '18 at 23:22
  • $\begingroup$ Again, by the Leibniz criterion. $\endgroup$ – José Carlos Santos Jul 13 '18 at 23:23

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