1
$\begingroup$

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?

  • 2
    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

2 Answers 2

4

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

  • 1
    @doniyor, it's never bad to clear one's doubt after sincere effort.2012-12-15
3

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

  • 0
    @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