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**Jhevon** ok, first of all, the power series for sine is defined for all real x, so we know the radius of convergence is $\displaystyle \infty$, and the interval of convergence is $\displaystyle (-\infty, \infty)$. However, if we are required to prove it, we can use the ratio test, beginning as follows:

We want $\displaystyle \lim_{n \to \infty} \left| \frac {a_{n + 1}}{a_n} \right| < 1$, where $\displaystyle a_n = \frac {\pi^{2n + 1} \cdot x^{2n + 1}}{(2n + 1)! \cdot 2^{2n + 1}}$, so we have

$\displaystyle \lim_{n \to \infty} \left| \frac {\pi^{2n + 3} \cdot x^{2n + 3}}{(2n + 3)! \cdot 2^{2n + 3}} \cdot \frac {(2n + 1)! \cdot 2^{2n + 1}}{\pi^{2n + 1} \cdot x^{2n + 1}} \right| $ $\displaystyle = \lim_{n \to \infty} \left| \frac {\pi^{2n + 3}}{\pi^{2n + 1}} \cdot \frac {(2n + 1)!}{(2n + 3)!} \cdot \frac {2^{2n + 1}}{2^{2n + 3}} \cdot \frac {x^{2n + 3}}{x^{2n + 1}} \right|$

now finish up