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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?

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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.2012-12-15

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$\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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    @doniyor, it's never bad to clear one's doubt after sincere effort.2012-12-15
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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 you2012-12-15
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    why are we taking the absolute value?2012-12-15
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    @doniyor because if a sequence converges absolutely then it converges.2012-12-15
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    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$2012-12-15
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    @F'OlaYinka oh you r right, this is the point i forgot.2012-12-15
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    @amWhy okay, i got it. great teachers! :)2012-12-15
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    @doniyor Does this makes sense now? Absolute convergence ==> convergence.2012-12-15
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    @amWhy i read once that if the absolute value of number converges, then it really converges, but now in our case, it doesnot importantly play role2012-12-15