Does the series $\sum_{n=2}^\infty \frac {\sin(n+\frac 1n)}{(\ln n)^2}$ converge? Could you give me some hint how to deal with this series ?
I could not conclude about absolute convergence because 
$\frac {\left|\sin\left( n+\frac 1n \right)\right|}{(\ln n)^2}\le \frac 1{(\ln n)^2}$
does not get me anywhere.
I tried to use the Dirichlet's test to prove convergence of this series:
$\sum_{n=2}^\infty \frac {\sin\left( n+\frac 1n \right)}{(\ln n)^2}$ 
but I could not prove that $\sum_{n=2}^\infty \sin(n+\frac 1n)$  are bounded.
Thanks.
 A: $\sin (n+\frac{1}{n}) = \sin n \cos \frac{1}{n} + \sin \frac{1}{n} \cos n$
Firstly $\sum_{n=2}^\infty \frac{\sin \frac{1}{n} \cos n}{\ln ^2 n}$ is absolutely convergent from comparison test, because
$\sum_{n=2}^\infty \left| \frac{\sin \frac{1}{n} \cos n}{\ln ^2 n} \right| \le \sum_{n=2}^\infty \frac{\frac{1}{n} \cdot 1}{\ln ^2 n} < \infty$
Convergence of last series follows for example from http://en.wikipedia.org/wiki/Cauchy%27s_condensation_test
Secondly observe, that function $f(x)=\frac{\cos \frac{1}{x}}{\ln ^2 x}$ is decraising for big $x$ because $f'(x)= \frac{\ln x \sin \left( \frac{1}{x} \right) -2x \cos \left( \frac{1}{x} \right)}{\ln ^3 x}$,
Easy to see that $\lim_{x\to \infty} f'(x) = -\infty$ therefore there is $M>0$ such that for $\mathbb{N} \ni n> M$ sequence $\left( f(n) \right)_{n>M}$ is decraising and tends to $0$. If we could show that $\left|\sum_{k=1}^n \sin k \right| $ is bounded, we would just apply Dirichlet test to show that $\sum_{n=2}^\infty \frac{\sin n \cdot \cos \frac{1}{n}}{\ln ^2 n}$ diverges.
It's well known that $\sum_{k=1}^n \sin ka = \frac{\sin \left(\frac{na}{2} \right) \sin \left( \frac{(n+1)a}{2} \right)}{\sin \left( \frac{a}{2}\right)}$ (you can prove it f.e by complex numbers), so 
$\left| \sum_{k=1}^n \sin k \right| \le \frac{1}{\left| \sin \frac{1}{2}\right|}$.
To sum up: $\sum_{n=2}^\infty \frac{\sin (n+ \frac{1}{n})}{\ln ^2 n}$ converges as sum of two convergent series.
A: *

*The series $\sum_{n\geqslant 2}\frac{\sin n}{(\log n)^2}$ is convergent (use a summation by parts argument). 

*We have $\left|\sin\left(n+\frac 1n\right)-\sin n\right|\leqslant \frac 1n$.

*The series $\sum_{n\geqslant 2}\frac 1{n(\log n)^2}$ is convergent. 
