Convergence of $\sum_{n = 1}^\infty \frac{\sin n (1-\sin n)}{n}$ How I can prove that this sum converges?
I can prove that  $\frac{\sin n}{n}$ converges and  that $(1-\sin n)$ is bounded.
 A: $$\sin^2 x  = \frac{1-\cos(2x)}{2}$$
then, 
$$\sum_{n=1}^{\infty} \frac{\sin n (1-\sin n)}{n} = \sum_{n=1}^{\infty} \left(\frac{\sin n}{n} - \frac{\sin^2 n}{n} \right)= \sum_{n=1}^{\infty} \left(\frac{\sin n}{n} + \frac{\cos(2n)}{2n}-\frac{1}{2n} \right).$$
But the series $$\sum_{n=1}^{\infty} \frac{\sin n}{n} $$
and $$\sum_{n=1}^{\infty}  \frac{\cos(2n)}{2n} $$ converges by Dirichlet test.
Whereas the series 
$$\sum_{n=1}^{\infty} \frac{1}{2n} $$ diverges.
Hence, $$\sum_{n=1}^{\infty} \left(\frac{\sin n}{n} + \frac{\cos(2n)}{2n}-\frac{1}{2n} \right)=\sum_{n=1}^{\infty} \frac{\sin n (1-\sin n)}{n}.$$ also diverges
A: The series is divergent. Note that 
$$\sum_{n=1}^{\infty} \frac{\sin n (1-\sin n)}{n} = \sum_{n=1}^{\infty} \left[ \frac{\sin n}{n} - \frac{\sin^2 n}{n} \right].$$
Now the series $\displaystyle \sum_{n=1}^{\infty} \frac{\sin n}{n}$ converges by the Dirichlet test. But the series $\displaystyle \sum_{n=1}^{\infty} \frac{\sin^2 n}{n}$ diverges to infinity, since for each $k \geqslant 1$ at least one of the numbers $n = 3k-2, 3k-1, 3k$ satisfies $|\sin n| \geqslant \frac{1}{2}$, so
$$\sum_{n=1}^{\infty} \frac{\sin^2 n}{n} \geqslant \sum_{k=1}^{\infty} \frac{\frac{1}{4}}{3k} = \infty.$$
Therefore
$$\sum_{n=1}^{\infty} \left[ \frac{\sin n}{n} - \frac{\sin^2 n}{n} \right] = -\infty.$$
A: The given series is not convergent by Kronecker's lemma, since the partial sums of $\sin(n)(1-\sin(n))$ are not bounded. We have $\left|\sum_{n=1}^{N}\sin(n)\right|\leq 2$ but 
$$ \sum_{n=1}^{N}\sin^2(n)\sim \frac{N}{2}. $$
