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'hospital's or dividing by the largest exponent from the denominator. here, i don't see why i would want to use l'hospital's, 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'hospital's or dividing by the largest exponent from the denominator. here, i don't see why i would want to use l'hospital's, 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.