Decide whether the series is absolutely convergent, conditionally convergent or divergent

Question

$$\sum_{n=1}^\infty(-1)^n\sin\left(\frac{a}{n}\right),\: a\in \mathbb{R}$$

I'm afraid I cannot use the alternating series test since $\sin(n)$ is not a decreasing sequence. Which test should I use instead?

Answer 1

Hint $:$ For sufficiently large $n,$ $\left | \sin \left (\frac a n \right ) \right |$ definitely decreases and converges to $0$. So use Leibnitz test for alternating series to complete. Since convergence or divergence of a series does not depend on finitely many terms of it you are through.

Observe that if $a=0$ the series $\sum\limits_{n=1}^{\infty} (-1)^n \sin \left ( {\frac a n} \right )$ is absolutely convergent and hence convergent as well.

Now for $a \neq 0$ we have $\lim\limits_{n \rightarrow \infty} \frac {\sin \left (\frac a n \right )} {\frac 1 n} = a \neq 0.$ Hence by the limit form of the comparison test it follows that the series $\sum\limits_{n=1}^{\infty} (-1)^n \sin \left ({\frac a n} \right )$ is not absolutely convergent (Since $\sum\limits_{n=1}^{\infty} \frac 1 n$ is divergent) though it is convergent. Therefore the series $\sum\limits_{n=1}^{\infty} (-1)^n \sin \left ( {\frac a n} \right )$ is conditionally convergent for $a \neq 0.$

Answer 2

Hint: compute the limit of $n\sin\frac{a}{n}$. After computing this limit, what does it tell you about the series $\sum_{n=1}^\infty \sin\frac{a}{n}$?