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? 

Notice that 
$$\sum_{k=1}^n\frac 1{(n+k)^2}=\frac 1{n}\frac 1n\sum_{k=1}^nf\left(\frac kn\right),$$
with $f(x):=\frac 1{1+x^2}$, a continuous positive function. Hence the convergence of the initial series reduces to the convergence of 
$$\sum_{n\geqslant 1}\frac{\sin^2(n\theta)}n.$$
Write $\sin^2(A)=\frac{1-\cos(2A)}2$. If $\theta$ is not a multiple of $\pi$, the series is divergent because $\sum_{n\geqslant 1}\frac{\sin(2n\theta)}n$ is convergent.
