Let $S_n=\sum_{i=1}^{n}\sin k,\quad S_0=0.$
Then
$\sum_{k=1}^{n}\frac{\sin k}{k}=\sum_{k=1}^{n}\frac{S_k-S_{k-1}}{k}=\frac{S_n}{n}+\sum_{k=1}^{n-1}\frac{S_k}{k(k+1)}.$
Could someone please explain the steps in getting this last equality. I am told it is using Abel's summation, but I have been reading up on that and I still can't see it.