Proving the series sin(n+1)-sin(n+2) converges or diverges As the title states, I'm trying to determine whether the series $$\sum_{n=1}^\infty \sin(n+1)-\sin(n+2)$$ converges or diverges. My intuition is saying diverging since the sines are oscillating. How will I go about 'formally' proving this? 

I searched online for this and I got an 'explanation' saying $s_n=\sin(2)-\sin(n+2)$. Where did this even come from, and how does this prove anything?
First: your sum diverges. To see this, write it out and note that it is a telescoping series- all but two of the terms cancel. Thats how you get your $s_n$. Then note that the value of the nth partial sum is actually a linear combination of $\sin(n)$, whose limit does not exist, therefore the limit of $s_n$ does not exist.
In order that $\sum_{n\geq 1}a_n$ is convergent, $\lim_{n\to +\infty}a_n = 0$ is needed.
On the other hand
$$\sin(n+1)-\sin(n+2) = -\cos\left(n+\frac{3}{2}\right)\cdot 2\sin\frac{1}{2} $$
and $\left\{\cos\left(n+\frac{3}{2}\right)\right\}_{n\geq 1}$ is dense in $(-1,1)$, hence $\limsup_{n\to +\infty}a_n = 2\sin\frac{1}{2}&gt;0$.

Proof of density:


$\pi $ is irrational
$\{e^{in}\}_{n\geq 1}$ and $\{e^{in+3i/2}\}_{n\geq 1}$ are dense in the unit circle (equidistributed, actually)
The projection $z\to\text{Re}(z)$ is a continuous map, hence it preserves density
$\left\{\cos\left(n+\frac{3}{2}\right)\right\}_{n\geq 1}$ is dense in $(-1,1)$.




We may also avoid to go through the irrationality of $\pi$ and the previous density argument. If we assume that $\sin(n+1)-\sin(n+2)$ has an absolute value smaller than $\varepsilon&gt;0$ for any $n\geq N_0$ we get a contradiction, since the function
$$ f(x)=\left|\sin(x+1)-\sin(x+2)\right|+\left|\sin(x+2)-\sin(x+3)\right|+\left|\sin(x+3)-\sin(x+4)\right| $$
is $\geq\frac{3}{2}$ for any $x\in\mathbb{R}$.



Summarizing, the sequence given by
$$ S_N = \sum_{n=1}^{N}\left[\sin(n+1)-\sin(n+2)\right]\stackrel{\text{telescopic sum}}{=}\sin(2)-\sin(N+2) $$
is not convergent.