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

Question

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?

Answer 1

Yes, it converges conditionally, by the Leibniz criterion.