$\lim_{n \to +\infty} n \cdot\cos^2\left(\frac{n \pi}{3}\right)$

Question

Find $\lim_{n \to +\infty} n \cdot \cos^2\left(\frac{n \pi}{3}\right)$

First I have a look at $\cos(\frac{n \pi}{3})$

What I expect is that the sequence diverges so I want to find two sub-sequences that tend to a different value, correct?

$\cos$ is periodic for $2 \pi$ so for the first sub-sequence let n:=6m and I get $\cos(\frac{6m \cdot \pi}{3})=\cos(2m \cdot \pi)$ which is 1 for every m.

For the next sub-sequence it seems to me that finding a k that gets me either $\cos(3\pi)$ or $\cos(\pi)$ would be a good idea since as then the sub-sequence will always deliver 0. However I get think of any k that satisfies that. Can someone help me out here? Thanks in advance!

Answer 1

We have that $\forall n\in \mathbb{N}$

$$\cos^2\left(\frac{n \pi}{3}\right) \ge \frac14$$

therefore

$$ n \cdot \cos^2\left(\frac{n \pi}{3}\right) \ge \frac n 4\to \infty$$

Answer 2

Unfortunately, you cannot find two subsequences convergent to different values since the general limit is $\infty$. Let $$a_{3n}=3n\cos^2{n\pi}=3n\\a_{3n+1}={(3n+1)}\cos^2{(n\pi +{\pi \over 3})}={3n+1\over 4}\\a_{3n-1}={(3n-1)}\cos^2{(n\pi -{\pi \over 3})}={3n+1\over 4}$$what can you conclude from here?