I have to show that the series $ {x_n} = { \sin (\log \ n ) }$ diverges.
If I start of with $ x_n - x_{n-1} $ that doesn't get me anywhere.
It is fairly obvious that $ \log n $ diverges. Can I use that and the fact that $ \sin x$ does not converge to say that the composition of these two series diverges as well?
In general, you cannot use that if a sequence $(a_n)_n$ diverges, the sequence $\sin(a_n)_n$ also diverges. Consider for example $a_n = 2πn$. Then $(a_n)_n$ is clearly divergent, but $\sin(a_n)_n$ is the zero sequence, which trivially converges.