I had to prove that the series $ \sum_{n=0}^{\infty} \cos^2(x)\sin^{2n}(x)$ is convergent for fixed x in $(-\frac{\pi}{2},\frac{\pi}{2})$, determine the sum, and prove, that the sum does not depend on x in $(-\frac{\pi}{2},\frac{\pi}{2})$
My original proof was as follows:Considering that $\cos^2$ does not depend on n, and can hence be considered as a constant. Which means I can rewrite the series as:$$ \cos^2(x) \sum_{n=0}^{\infty} \sin^{2n}(x)$$
The series $\sum_{n=0}^{\infty} \sin^{2n}(x)$ can be written as
$$ \sum_{n=0}^{\infty} \sin^{2n}(x) = sin^{2\cdot0}(x)+sin^{2\cdot 1}(x)+...sin^{2n}(x) $$
$$\sum_{n=0}^{\infty} \sin^{2n}(x) = 1+ \sin^{2}(x)(1+\sin^{2}(x)+ \sin^{4}(x)...$$
Which I further rewrite:
$$\sum_{n=0}^{\infty} \sin^{2n}(x) = 1+ \sin^{2}(x)(\sum_{n=0}^{\infty} \sin^{2n}(x))$$
Which can be simplified too
$$\sum_{n=0}^{\infty} \sin^{2n}(x)- \sin^{2}(x) (\sum_{n=0}^{\infty}\sin^{2n}(x)) = 1$$
$$\sum_{n=0}^{\infty} \sin^{2n}(x)(1- \sin^{2}(x)) = 1$$
$$\sum_{n=0}^{\infty} \sin^{2n}(x)= \frac{1}{1-\sin^{2}(x)}$$$$\sum_{n=0}^{\infty} \sin^{2n}(x)= \frac{1}{\cos^{2}(x)}$$
Which means that I can rewrite the original equation$$ \cos^2(x)\sum_{n=0}^{\infty} \sin^{2n}(x)= \frac{1}{\cos^{2}(x)} \Rightarrow \cos^2(x)\frac{1}{\cos^{2}(x)}= 1 $$
Which means that $ \sum_{n=0}^{\infty} \cos^2(x)\sin^{2n}(x)$ is convergent for fixedx in $(-\frac{\pi}{2},\frac{\pi}{2})$, and the sum is equal to 1. I have also proven that the sum does not depend on x in $(-\frac{\pi}{2},\frac{\pi}{2})$, and only the value of n
My teacher told my solution was entirely incorrect and that I would completely start over. However, a commenter on this board confirmed that my solution was correct. So, my question is, should I ask to get my assignment regraded?
EDIT: NEW PROOF
Given the series $ \sum_{n=0}^{\infty} \cos^2(x)\sin^{2n}(x)$, for $x \in (-\frac{\pi}{2},\frac{\pi}{2})$ both $\cos^2(x)$ and $\sin^2(x)$ are bounded between 0 and 1. So, $\cos^2(x)\sin^{2n}(x)$≤$\sin^{2n}(x)$ which is a geometric series with a common ratio r = $\sin^{2}(x)$
Since 0<$\sin^{2}(x)<1$ for x $\in (-\frac{\pi}{2},\frac{\pi}{2})$, the geometric series $ \sum_{n=0}^{\infty} \sin^{2n}(x)$ converges absolutely.
I find the sum using the formula of infinite geometric series, $S=\frac{a}{1-r}$, where a=$\cos^2(x)$ and r= $sin^2(x)$$$S= \frac{\cos^2(x)}{1-\sin^2(x)} =\frac{\cos^2(x)}{\cos^2(x)} = 1 $$
Thus the sum is also independent of x $\in (-\frac{\pi}{2},\frac{\pi}{2})$