Does the series $\sum_{n=2}^{\infty} \frac {\sin(n+\frac{1}{n})}{\log\log n}$ converge?

Question

Does the series $\sum_{n=2}^{\infty} \frac {\sin(n+\frac{1}{n})}{\log\log n}$ converge?

My attempts :

$\sin(x+y) = \sin x\cos y + \cos x \sin y$

now $\sin(n +\frac{1}{n}) = \sin n \cos \frac{1}{n} + \cos n\sin\frac{1}{n}$

now $\sum_{n=2}^{\infty} \frac {\sin n \cos 1/n }{\log\log n} + \sum_{n=2}^{\infty}{\frac {\cos n \sin 1/n}{\log\log n}}$

After that I can not able to proceed further.

Answer 1

\begin{align} \sum_{n=2}^{\infty} \frac {\sin( n + 1/n) }{\log\log n}&=\sum_{n=2}^{\infty} \frac {\sin n \cos 1/n }{\log\log n} + \sum_{n=2}^{\infty}{\frac {\cos n \sin 1/n}{\log\log n}}\\ &=\sum_{n=2}^{\infty} \sin n\,\frac {1 }{\log\log n} +\sum_{n=2}^{\infty} \frac {(\sin n)(\cos 1/n-1) }{\log\log n}+ \sum_{n=2}^{\infty}{\cos n\,\frac { \sin 1/n}{\log\log n}}. \end{align} By Dirichlet's test, the first and last series converge. The second one is absolutely convergent, since $\cos 1/n-1\sim -1/(2n^2)$.

Answer 2

Use Dirichlet's test in order to show that it converges.