Is the Series : $\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + ..........+\frac 1 {(n+n)^2} \right) \sin^2 n\theta $ convergent? 
Is the Series :
$$\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \sin^2 n\theta $$ convergent?


Attempt: 
$$\sum_{n=0}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \sin^2 n\theta $$
$$= \sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) (1- \cos( 2n\theta)) \frac 1 2$$
$$=\frac 1 2 \left[\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right)\right]- \frac 1 2\left[\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \cos(2n\theta)\right] $$ 
The left part by sandwich theorem has limiting value to $0$
The right part by Dirichlets theorem is convergent as $  \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right)$ is a positive, monotonically decreasing sequence and the sequence of partial sum of $\sum_{n=1}^{\infty} \cos 2n\theta$ is bounded.
Hene, the given series is convergent. Am i correct? 

if $\theta=k\pi$, the series is  convergent.
If $\theta$ is not a multiple of $\pi$, 
$$\left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \sin^2 n\theta \geq \frac n{(n+n)^2}  \sin^2 n\theta =\frac {\sin^2 n\theta}{4n} =\frac1{8n}- \frac{\cos 2n\theta}{8n}$$
$\sum \frac{\cos 2n\theta}{8n} $ is convergent,  hence,  the series is divergent
