Does $$\sum_{n=1}^\infty(-1)^n \sin \left( \frac{1}{n} \right)
$$ converge conditionally or absolutely?
I know that this series converges conditionally using the Leibniz's convergence test, but what method should be used to decide whether it converges absolutely?
Best Answer
No, it does not converge absolutely. Note $\sin x \sim x$ for small $x$ and hence $\sin \frac 1n \sim \frac 1n$ for large $n$. This implies $\sin \frac{1}{n} \geq \frac{1}{2n} \geq 0$ for large $n$. But we know that $\sum \frac{1}{2n} = \frac{1}{2} \sum \frac{1}{n} =+\infty$ and so by comparison $\sum \sin \frac{1}{n} = + \infty$.
However, the series converges conditionally. This is an immediate consequence of the alternating series test. $\left|\sin \frac{1}{n}\right| = \sin \frac{1}{n} \to 0$ as $n \to \infty$ and $\sin \frac{1}{n}$ is positive and monotonically decreasing.