Answer
$$\rho = \infty$$
Work Step by Step
1. Find the Taylor series:
$$sin(x) = sin(0) + cos(0)x - \frac{sin(0)}{2!}x^2 - \frac{cos(0)}{3!}x^3 + \frac{sin(0)x^4}{4!} + \frac{cos(0)x^5}{5!}...$$ $$sin(x) = 0 + x - 0 - \frac{x^3}{3!} + 0 + \frac{x^5}{5!}...$$ $$sin(x) = \sum^{\infty}_{n = 0} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$
2. Use the ratio test to test for convergence:
$$\lim_{n \longrightarrow \infty}
\Bigg| \frac{ (-1)^{n+1} \frac{x^{2(n+1)+1}}{(2(n+1)+1)!} } { (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} } \Bigg |
= \lim_{n \longrightarrow \infty} \Bigg| (-1) \frac{x^2}{(2n+2)(2n+3)} \Bigg| = 0$$
- Therefore, the series converges absolutely when $0 \lt 1$, which is true for any $x$ value.
$$\rho = \infty$$
