# Is this series conditionally convergent or absolutely convergent? $\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\sin\left(\frac{1}{k}\right)$

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?

• 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. – Lithium Jul 13 '18 at 23:10
• 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. – José Carlos Santos Jul 13 '18 at 23:13
• May I ask you that why this one is conditionally convergent? $\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{k^{k}}{\left(k+1\right)^{k+1}}$ – Lithium Jul 13 '18 at 23:22