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

Question

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

My solution:

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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*.

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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 $n$.

Answer 1

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.

Answer 2

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.)