solve $\lim_{n \to \infty} \frac{(-2)^n}{3^{2n}} $

Question

i am eating myself not being able to solve this problem. i somehow feel that the sequence converges to $0$, but once i calculate, it is not coming to that result. or am i making stupid mistake on the way?

my steps:

$$\lim_{n \to \infty} \frac{(-2)^n}{3^{2n}} = \frac{\lim_{n \to \infty} (-2)^n }{\lim_{n \to \infty} 3^{2n} } = \frac{diverging}{diverging} = ? $$

can someone please help me?

Answer 1

$\lim_{n \to \infty} \frac{(-2)^n}{3^{2n}} = \lim_{n \to \infty} \left(\frac{-2}{3^2}\right)^n=0 $ as $\mid \left(\frac{-2}{3^2}\right)\mid<1$

Answer 2

Note that:

$$\lim_{n \to \infty} \frac{(-2)^n}{3^{2n}} = \lim_{n \to \infty} \frac{(-2)^n}{(3^2)^n} = \lim_{n \to \infty} \left(\frac{-2}{3^2}\right)^n$$

Note that $\Big|\dfrac{-2}{3^2}\Big| = \dfrac{2}{9}< 1$, so we have $$\lim_{n \to \infty} \left(\frac{-2}{3^2}\right)^n = 0.$$