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

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:

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

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Note that $\lim \cfrac{a_n}{b_n}$ is defined whereas $\cfrac{\lim a_n}{\lim b_n}$ is not in your case. The implication is the other way around. If you have two convergent sequences, the sequence of the quotients is the quotient of the limits. But having a converging quotient doesn't imply the numerator and the denominator both converge. – xavierm02 Dec 15 '12 at 16:10

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

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thanks @lab, sorry for the bad question – doniyor Dec 15 '12 at 16:07
@doniyor, it's never bad to clear one's doubt after sincere effort. – lab bhattacharjee Dec 15 '12 at 16:09

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 $$\lim_{n \to \infty} \left(\frac{-2}{3^2}\right)^n = 0.$$

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yes, i was just overthinking, thank you – doniyor Dec 15 '12 at 16:09
why are we taking the absolute value? – doniyor Dec 15 '12 at 16:21
@doniyor because if a sequence converges absolutely then it converges. – user31280 Dec 15 '12 at 16:22
There really is no need: just recognize the absolute value of a number $p\over q$ is less than 1, then the limit as n goes to infintity of $(\frac{p}{q})^n = 0$ – amWhy Dec 15 '12 at 16:23
@amWhy okay, i got it. great teachers! :) – doniyor Dec 15 '12 at 16:25