# Convergence of a sequence with both sin and cos

I'm trying to figure out whether the following series converges absolutely or conditionally or whether it diverges. I am stuck on the following one that involved both sin and cosine: $$\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2+\cos(n)}$$ Any guidance would be greatly appreciated however :)

• Perhaps you can check whether it converges absolutely or not. – MathNewbie Jun 11 '15 at 23:07
• Also, as a friendly psa, you can check out the link here to learn how to properly type mathematics on this site to make your equations readable. – JMoravitz Jun 11 '15 at 23:11
• That's what I'm asked to do but I'm confused as how to start :) – Chloe Jun 11 '15 at 23:11
• Hint: Try and use the fact that $-1\leq\sin(n)\leq 1$. Similar for $\cos(n)$. – OnceUponACrinoid Jun 11 '15 at 23:12
• You can use Limit Comparison with $\sum_1^\infty \frac{1}{n^2}$. Or else use Comparison with $\sum_2^\infty \frac{1}{n^2/2}$. – André Nicolas Jun 11 '15 at 23:18

$$\left|\frac{\sin n}{n^2+\cos n}\right|\le \frac{1}{n^2-1}$$
for $n>1$.
• Technically you also have to verify $\cos 1 \neq -1$ even if you accept $\cos n \geq -1$, otherwise the series might blow up with the first term and not be defined. – user2566092 Jun 11 '15 at 23:21
• @user2566092 this didn't claim to be a full solution, and checking that $\cos 1\neq -1$ is a trivial matter. – JMoravitz Jun 11 '15 at 23:23
• $\frac{1}{n^2-1} < \frac{1}{n^2}$ is not necessarily true. But you can invoke the limit comparison test between the two series $\sum \frac{1}{n^2-1}$ and $\sum \frac{1}{n^2}$. The p-series test and limit comparison test tells us that $\sum \frac{1}{n^2-1}$ must converge. – MathNewbie Jun 11 '15 at 23:31