# convergence of $\sum_{n}^{\infty}(-1)^n\sin \frac{2}{n}\cos \frac{1}{n}$

I have problem similar to this convergence of $\sum_{n}^{\infty}(-1)^n \log(1+\sin(\sqrt{n+1}-\sqrt n)$ . With series $\sum_{n}^{\infty}(-1)^n\sin \frac{2}{n}\cos \frac{1}{n}$. I want to show that it is not increasing. I know that $\cos x$ is increasing as it approaches 0 and $\sin x$ is decreasing as it approaches 0. I think I need to show somehow that sin decrease faster than cos. But how do I do that? I thought about deriving and seeing if it's smaller than zero, but I don't think that it is good way.

$\sin\frac{2}{n} \cos\frac{1}{n} =2\sin\frac{1}{n} \cos^2 \frac{1}{n} =2\sin\frac{1}{n} -2\sin^3 \frac{1}{n}$
• it is: $\sin(2x) = 2\sin(x)\cos(x)$ and then $\cos^2(x) + \sin^2(x)=1$ – mookid Jun 16 '14 at 15:34
• I used the identity $\sin (2 x) =2\sin x \cos x$ and $\cos^2 x =1-\sin^2 x.$ – user110661 Jun 16 '14 at 15:34