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