Does the series of cosine converge or diverge? Does the series of cosine 
$$\sum\limits_{n=1}^\infty\frac{\cos (\pi n)}{n}$$ 
converge or diverge?
 A: The series converges, in fact we have
$$
\sum_{n=1}^\infty\frac{\cos\pi n}{n}=\sum_{n=1}^\infty\frac{(-1)^n}{n}=-\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=-\ln2.
$$
A: It's an alternating series where the numerator is $-1$ or $1$. Denominator is linear increasing. It meets the prereqs to be conditionally convergent.
A: It converges.
$$\sum_{n=1}^{\infty} \frac{\cos \left(\pi n\right) }{n}
 =-\frac{1}{1}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\dots+\left(-1\right)^k \frac{1}{k} 
 = -\sum_{n=1}^{\infty} \frac{1}{2n-1}+\sum_{n=1}^{\infty}\frac{1}{2n}
 =\sum_{n=1}^{\infty} \left(-\frac{1}{2n-1}+\frac{1}{2n}\right) $$
By Integral test for convergence: 
$\sum_{n=N}^{\infty} f(n)$ converges if $\int_{N}^{\infty}f(x)dx$ converges. 
$\sum_{n=N}^{\infty} f(n)$ diverges if $\int_{N}^{\infty}f(x)dx$ diverges.
$$ \int_{1}^{\infty}\left(\frac{1}{2x}-\frac{1}{2x-1}\right)dx=\frac{1}{2}\ln\left|\frac{2x}{2x-1}\right|\bigg|_1^\infty=\frac{1}{2}\left(\ln{2}-\ln{1}\right)=\frac{\ln 2}{2}$$
The series converges because the integral converges. You can check other convergences test like the d'Alamber test.
