Limit of (cos(n))^2 / 2^n Sequence - Homework Help

[arl146]

Homework Statement

the sequence is a_n = (cos(n))^2 / 2^n

Homework Equations

none really

The Attempt at a Solution

like i mentioned in my last post, i usually use l'hopitals or dividing by the largest exponent from the denominator. here, i don't see why i would want to use l'hopitals, so that's out of the question. i can't really divide by the largest exponent. is there a way i could break this down or something?

 

[Curious3141]

Homework Statement

the sequence is a_n = (cos(n))^2 / 2^n

Homework Equations

none really

The Attempt at a Solution

like i mentioned in my last post, i usually use l'hopitals or dividing by the largest exponent from the denominator. here, i don't see why i would want to use l'hopitals, so that's out of the question. i can't really divide by the largest exponent. is there a way i could break this down or something?

What's the maximum value the numerator can take?

 

[arl146]

1. so that means the bottom is always going to be larger.
can i just write on my homework like ..

(cos(n))^2 < 1, for all n
as n approaches infinity, 2^n gets large.
small # / large # = 0

 

[Curious3141]

1. so that means the bottom is always going to be larger.
can i just write on my homework like ..

(cos(n))^2 < 1, for all n
as n approaches infinity, 2^n gets large.
small # / large # = 0

Yes, although I would write \cos^2 n \leq 1 \forall n \in \mathbb{N}. The strict inequality is actually true because the only values of n that give you a value of exactly one are zero and multiples of \pi and you're dealing with natural numbers here, but proving it is not worth the trouble, and is unnecessary here.

So just write that, and \lim_{n \rightarrow \infty} 2^n = \infty, so the quotient tends to zero.