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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.

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    $\begingroup$ Are you guaranteed that $|\sin x| < 1$ always in this context? Because $\sin(\pm \pi/2) = \pm 1$, for example, and in those cases the series won't converge. $\endgroup$ – tilper Feb 19 '18 at 17:36
  • $\begingroup$ The series does not always converge. It diverges if $x = \frac{\pi}{2}$ $\endgroup$ – mephistolotl Feb 19 '18 at 17:37
  • $\begingroup$ It is also not clear whether it is $\sin x^n$ or $(\sin x)^n$ (title vs body of the question) $\endgroup$ – amrsa Feb 19 '18 at 17:39
  • $\begingroup$ For |sinx|<1 replace sinx with m. Now the sequence turns into $n.m^n$ which is simple A.G.P $\endgroup$ – NewGuy Feb 19 '18 at 17:39
  • $\begingroup$ Hint: this is of the form $\sum_{n=0}^\infty ny^n = y\sum_{n=0}^\infty ny^{n-1}$. Can you find the sum of the latter series? BTW $|\sin x| \le 1$: it's not always less than 1. $\endgroup$ – NickD Feb 19 '18 at 17:44
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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).

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  • $\begingroup$ Not sure you can interchange the limit of the derivative with the limit of the sum. Don't you have to have uniform convergence or something like that? $\endgroup$ – Adrian Keister Feb 19 '18 at 17:40
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    $\begingroup$ @AdrianKeister The classical theorem is that if each $f_n$ and $f_n'$ are continuous on an interval, the series $$ f(x) = \sum_{n=1}^\infty f_n(x) $$ converges pointwise, and the series $$ g(x) = \sum_{n=1}^\infty f_n'(x) $$ converges uniformly, then $f$ is differentiable and $f' = g$. $\endgroup$ – Von Neumann Feb 19 '18 at 17:45

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