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: One does not need to appeal to Dirichlet's test or resort to calculus to determine convergence.  Rather, we simply note that we can write
$$\begin{align}
\sum_{k=1}^K \frac{\cos(k)-\cos(k+1)}{k}&=\sum_{k=1}^K \frac{\cos(k)}{k}-\sum_{k=1}^K \frac{\cos(k+1)}{k}\\\\
&=\sum_{k=1}^K \frac{\cos(k)}{k}-\sum_{k=2}^{K+1} \frac{\cos(k)}{k-1}\\\\
&=\cos(1)-\sum_{k=2}^K \frac{\cos(k)}{k(k-1)}-\frac{\cos(K+1)}{K}\tag1
\end{align}$$
The second term on the right-hand side of $(1)$ converges as the denominator is $O(1/k^2)$ (in fact, the sum is bounded in absolute value by $1$) and the third term tends to $0$ as $K\to\infty$.  Hence, the series converges as was to be shown!
A: Hint- $\operatorname{cos}(k)-\operatorname{cos}(k+1)=2\operatorname{sin}\frac{2k+1}{2}\operatorname{sin}\frac{1}{2}$
A: This is not what was asked for in the question, but it is interesting enough that I thought I would post an answer about it.
Let's write the first few terms. $$\cos(1)-\cos(2)+\frac{\cos(2)}{2}-\frac{\cos(3)}{2}+\frac{\cos(3)}{3}-\frac{\cos(4)}{3}+...$$
Let $a_n$ be the coefficient in front of $\cos(n)$. So then our sum is $$\sum_{n=1}^{\infty}{a_n\cos(n)}$$
We can see that $a_1=1$, $a_2=-1+\frac{1}{2}$, $a_3=-\frac{1}{2}+\frac{1}{3},...$
And in general, $a_n=\frac{-1}{n-1}+\frac{1}{n}$ for $n \neq1$ and with $a_1=1.$ This sum can be thought of a Fourier series of some function $f$ with a period of $2\pi$ evaluated at $x=1$ with satisfying $a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}{f(x)}\mathrm{d}x=0$ and $b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)\mathrm{d}x=0 \ \forall n \in \mathbb{N}$. I think it would be very interesting if someone could find such a function $f$. 
