Prove that $\sum_{n=1}^{\infty}{\sin(\frac{\pi}{n})\cos({\pi n})}$ is convergent, but not absolutely convergent.

Prove that $$\sum_{n=1}^{\infty}{\sin(\frac{\pi}{n})\cos({\pi n})}$$ is convergent, but not absolutely convergent.

My solution:

Proof of convergence:

Observe that $$\cos(n \pi) = (-1)^{n}$$

so $$\sum_{n=1}^{\infty}{\sin(\frac{\pi}{n})(-1)^{n}}$$ but this is convergent*.

Proof of absolute convergence:

$$\sum_{n=1}^{\infty}{|\sin(\frac{\pi}{n})(-1)^{n}|} > \sum_{n=2}^{\infty}{\sin(\frac{\pi}{n})} > \sum_{n=2}^{\infty}\frac{1}{n} = \infty$$

Using fact that $$\sin(x) > \frac{x}{\pi}$$ for $$x \in [0, \pi/2]$$

Please check my solution and I need suggestion how should proof*. Intuition tell me that it converges, because we have the alternating sign $$(-1)^{n}$$, so we add and subtract terms for even and odd $$n$$.

• For * use alternating series test. Commented Jul 20, 2019 at 9:55
• But we have $\frac{1}{n} > \sin\left(\frac{\pi}{n}\right)$ for $n = 1$. Commented Jul 20, 2019 at 10:25
• @ViktorGlombik of course, I checked this gap. Commented Jul 21, 2019 at 11:09

Hints. For * use alternating series test and the fact that $$\sin{x}$$ is ascending/increasing on $$\left[0,\frac{\pi}{2}\right]$$. First of all $$\lim\limits_{n\to\infty}\sin{\frac{\pi}{n}}= \lim\limits_{n\to\infty}\frac{\pi}{n} \cdot \frac{\sin{\frac{\pi}{n}}}{\frac{\pi}{n}} \rightarrow 0$$ because $$\lim\limits_{\ x\to 0}\frac{\sin{x}}{x}=1$$. And $$n>m \geq2 \Rightarrow 0<\frac{\pi}{n}<\frac{\pi}{m}\leq \frac{\pi}{2} \Rightarrow 0<\sin{\frac{\pi}{n}}<\sin{\frac{\pi}{m}}<1$$ As a result, for $$a_n={\sin(\frac{\pi}{n})(-1)^{n}}$$ we have $$\lim\limits_{n\to\infty}a_n=0$$ and $$|a_n|$$ is monotonically decreasing.
The real fact is that $$\sin(x) \leq x$$ for $$x\in [0, \pi/2]$$. But, I think it is not hard to find an appropriate $$c$$ such that $$\sin(x) \geq cx$$ for $$x\in [0, \pi/2]$$. (Try to draw a graph and look at it seriously.)
• I compute $c= 1/\pi$ Commented Jul 20, 2019 at 9:16
• @MartinInf1n1ty I think that works, and maybe the optimal one is $c = 2/\pi$. Commented Jul 20, 2019 at 9:35
• @MartinInf1n1ty. Since $\lim_{n\to \infty}\frac {\sin \pi/n}{\pi/n}=1,$ we have $\sin (\pi/n)>\frac {1}{2}(\pi/n)$ for all but finitely many $n\in \Bbb N.$ Commented Jul 20, 2019 at 10:02