How to show that the serie

$$\sum \frac{\cos n -\sin n\pi}{n}$$


In general, I am having troubles with the series that include trigonometric functions, but I think there is no general approach to find if it converges or not.

I would be glad if I could have some help.

Thanks !


Since $\sin{n\pi}=0$, we only consider $\sum_{n=1}^\infty \frac{\cos{n}}{n}$. By means of Dirichlet test, since $1/n\to 0$ monotonically and $$\sum_{k=1}^n \cos{k}=\frac{1}{\sin{\frac{1}{2}}}\sum_{k=1}^n \cos{k}\sin{\frac{1}{2}}=\frac{1}{2\sin{\frac{1}{2}}}\sum_{k=1}^n(\sin{(k+1/2)}-\sin{(k-1/2)})=\frac{\sin{(n+1/2)}-\sin{1/2}}{2\sin{1/2}}$$ is bounded. We finish the proof.


$\sin k \pi=0$ for all $k \in \mathbb{Z}$, so you only got $\sum_{k=1}^{\infty} \frac{\cos k}{k}$ which converges

EDIT: use Dirichlet's test. $\sum_{k=1}^{n}\cos k <2$ and $\lim_{k \to \infty} \frac{1}{k} = 0$.

  • $\begingroup$ I know, but how to prove that this serie convergea? $\endgroup$ – Giiovanna Aug 27 '14 at 10:06
  • $\begingroup$ please see the edit $\endgroup$ – Alex Aug 27 '14 at 10:09

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