I've been struggling with this problem. I'm asked to verify where this series converges:
$$\sum_{n=0}^\infty n\cdot(\sin x)^n$$
and to find the sum of the series.
Attempted Solution:
With the ratio test we find that the series always converges, since $\vert \sin x \vert<1$ always. But I have no idea how to find the sum of this series.
Notice that
$$n\sin^n(x) = \frac{\sin(x)}{\cos(x)}\frac{d}{dx}\sin^n(x)$$
hence
$$\frac{\sin(x)}{\cos(x)}\frac{d}{dx} \sum_{n = 0}^{+\infty} \sin^n(x) = \frac{\sin(x)}{\cos(x)}\frac{d}{dx}\left(\frac{1}{1-\sin (x)}\right)$$
The sum has been summed with the Geometric Series.
Now take the derivative and arrange:
$$\frac{\sin(x)}{\cos(x)} \frac{\cos (x)}{(1-\sin (x))^2} = \color{red}{\frac{\sin(x)}{(1 - \sin(x))^2}}$$
Which is the sum.
P.s. It clearly diverges for $x = \pi/2$ which is a value (plus the relatives) that cannot be counted (it also make the geometric series to fail).