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

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

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.

• Are you considering $n \to 0$ or $n \to \infty$? When you say series are you considering the sum? – Henry Feb 22 '20 at 13:21
• Worth noting: the series (or sequence) is not alternating. Yes, there are both positive and negative terms, but the pattern of signs is tricky. Leibniz does not apply. – lulu Feb 22 '20 at 13:24
• @Henry The limit goes to infinity and no I am not considering the sum... I think. All the exercise asks is to find whether it converges or not. – Segmentation fault Feb 22 '20 at 13:27
• The word series has a definite meaning. Are you considering the sequence of terms $\sin n/n^2$? If so this goes to $0$ at $n=+\infty$ since the numerator is bounded by a constant and the denominator becomes arbitrarily large. – Allawonder Feb 22 '20 at 13:29
• You have presented a sequence, but mention an alternating series. Are you looking at $\lim\limits_{n\to\infty}\frac{\sin(n)}{n^2}$ or $\sum\limits_{n=1}^\infty\frac{\sin(n)}{n^2}$? – robjohn Feb 22 '20 at 14:01

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