Find whether $a_n = \frac{\sin(n)}{n^2}$ converges

Question

$a_n = \frac{\sin(n)}{n^2}$

Because this is an alternating series first I tried to find whether $|a_n|$ converges.

$$|a_n|= \frac{|\sin(n)|}{n^2}$$

I tried to compare this with $1/n^2$:

$$\lim \frac{\frac{|\sin(n)|}{n^2}}{\frac{1}{n^2}} = \lim |\sin(n)|$$

I'm unsure about what to do next? This limit goes anywhere between $0$ and $1$. Since $1/n^2$ converges, so does this series. Because when the limit is $]0;1]$ they both converge, and when it's 0 since $1/n^2$ converges, so does $a_n$. Is this correct? If so, then Leibniz's criteria isn't applied here and the series is absolutely convergent.

Answer 1

I think you can easily show that this sequence converges absolutely which implies convergence of sequence.

To show absolute convergence of sequence, \begin{align*} |a_n| &= \frac{|\sin(n)|}{n^2}\\ &\leq \frac{1}{n^2} \to 0 \text{ as } n \to \infty \end{align*}

To show convergence of series, we first show convergence of $\sum |a_n|$.

\begin{align*} \sum_{n=1}^\infty |a_n| &= \sum_{n=1}^\infty \frac{|\sin(n)|}{n^2}\\ &\leq \sum_{n=1}^\infty \frac{1}{n^2}\\ &= \frac{\pi^2}{6} \end{align*} This implies for all $N \in \mathbb{N}$, we have $$\sum_{n=1}^N |a_n| \leq \frac{\pi^2}{6}$$ Since the partial sums $\sum \limits_{n=1}^N |a_n|$ are upper bounded, Theorem 1 implies $\sum |a_n|$ indeed converges.

Theorem 1 A series of nonnegative terms converges if and only if its partial sums form a bounded sequence.

Convergence of $\sum |a_n|$ implies convergence of $\sum a_n$