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?
 A: 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.
A: As others have noted, the $N$th partial sum of the series in this problem is $\sin 2 - \sin (N+2).$ So to show the series diverges, it's enough to show $\sin n,n=1,2,\dots$ diverges.
We can prove this by noting that $e^{in}$ lands in the arc $A=\{e^{it}: \pi/4 < t < 3\pi/4\}$ infinitely many times. Thus $\sin n > 1/\sqrt 2$ for infinitely many $n.$ Similarly, $\sin n <- 1/\sqrt 2$ for infinitely many $n.$ It follows that $\sin n$ diverges.
Why must we have $e^{in}\in A$ infinitely many times? As $n$ goes through $1,2,\dots ,$ $e^{in}$ marches clockwise around the unit circle in steps of arc length $1.$ The arc $A$ has arc length $\pi/2 >1.$ Once every full orbit of the circle, $e^{in}$ has to land in $A,$ since it can't "jump over" an arc of length $>1$ by taking steps of length $1.$ So in fact $e^{in}\in A$ at least once in every $7$ consecutive values of $n$ ($7>2\pi !$).  
