I've calculated the radius of convergence for $\displaystyle \sum_{n =0}^{\infty} \sin^{2n} x $ in two different ways and I get two different answers.Find the range of values of values of x for which the series $\displaystyle \sum_{n=0}^{\infty} \sin^{2n} (x)$ and $\displaystyle \sum_{n=0}^{\infty} 2n \sin ^{2n-1}(x)$ converge, and find an expression for their sums, carefully justifying your answers.

The first way:

Let $\displaystyle a_n= \sin^{2n} x$.

Therefore $\displaystyle limsup_{n \rightarrow \infty} | \sin^{2n} x|^{\frac{1}{n} }=limsup_{n \rightarrow \infty} | \sin^2 x|=1$

Hence $\displaystyle R=\frac{1}{1}=1$.

Alternatively, using the ratio test:

$\displaystyle \left| \frac{a_{n+1}}{a_n} \right|=\left|\frac{\sin^{2n+2} x}{ \sin^{2n} x} \right|=\left|\sin^2 (x) \right|< 1 \Leftrightarrow \sin x<1 \Rightarrow x<\frac{\pi}{2}$

This tells me that $\displaystyle R=\frac{\pi}{2}$.

Which one is correct.

For $\displaystyle \sum_{n=0}^{\infty} 2n \sin^{2n-1} x$:

$\displaystyle \left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{2(n+1) \sin^{2n+1} x}{2n \sin^{2n-1} x } \right|= \left| \frac{n+1}{n} \sin^2 x \right| \rightarrow |sin^2 x|<1 \Rightarrow \ x< \frac{\pi}{2}$

The main problem I have now is trying to find the expression for the sums. Can someone show me how to do that?