Q:

Does $\sum \limits_{k=1}^{\infty} \frac{\cos(k)-\cos(k+1)}{k}$ converge?

I was given the following series and I'm asked to decide (and prove) whether it converges or diverge:
$\displaystyle\sum \limits_{k=1}^{\infty} \frac{\cos(k)-\cos(k+1)}{k}$
So far, I couldn't successfully approach the solution. At least, I've tried to use series expansion for $\cos$ and then received
$$\displaystyle\sum \limits_{k=1}^{\infty} \frac{\cos(k)-\cos(k+1)}{k} = \sum \limits_{k=1}^{\infty} \frac{\sum \limits_{n=0}^{\infty} \frac{(-1)^n\cdot k^{2n}-(-1)^n\cdot(k+1)^{2n}}{2n!}}{k} =$$
$$\sum \limits_{k=1}^{\infty} \frac{0}{k}+\sum \limits_{k=1}^{\infty} \frac{\sum \limits_{n=1}^{\infty} 
\frac{(-1)^n\cdot k^{2n}-(-1)^n\cdot(k+1)^{2n}}{2n!}}{k} = \sum \limits_{k=1}^{\infty} \frac{\sum \limits_{n=1}^{\infty} \frac{(-1)^n\cdot k^{2n}-(-1)^n\cdot(k+1)^{2n}}{2n!}}{k}$$
Can someone give me a hint to go on or can provide a different type of a possible approach to the solution?
Thanks in advance

A:

Hint- $\operatorname{cos}(k)-\operatorname{cos}(k+1)=2\operatorname{sin}\frac{2k+1}{2}\operatorname{sin}\frac{1}{2}$

