Uniform convergence of $\sum \cos^n x \sin x$ on $[0, \pi]$?

Question

I wonder how to check uniform convergence on the closed interval $[0, \pi]$ of the following series: $$ \sum_{n=1}^{\infty} \cos^n x \sin x $$ I came to consider this through a modification of a simpler example $\sum n\cos^n x \sin x$ where the necessary condition for uniform convergence is not met, i.e. the sequence $n\cos^n x \sin x$ does not converge uniformly to $0$ (as demonstrated here). However, in the modified example, this line of argument does not seem to work. How would you prove/disprove uniform convergence in this case?

Answer 1

You can compute this sum S. $$S(0) = S(\pi) = 0$$ And, if $0 <x< \pi$ :

$$S(x) := \sum_{n=1}^{\infty} \cos^n(x) \sin(x) = \sin(x) \cos(x) \sum_{n=0}^{\infty} \cos^n(x) = \frac{1}2 \frac{\sin(2x)}{1 - \cos(x)}$$

The last expression diverges when $x \to 0$. Indeed, with $\epsilon_i(x) \to 0$ when $x \to 0$

$$\frac{\sin(2x)}{1 - \cos(x)} = \frac{2x + x \epsilon_1(x)}{x^2/2+ x^2 \epsilon_2(x)/2} = 4 \frac{2 + \epsilon_1(x)}{x + x\epsilon_2(x)} \displaystyle\to +\infty \quad [x \to 0^{+}]$$

Thus, there is still a problem at $x = 0$.