sequences with cosines are they always divergent?

Question

I have to show if this sequence is converging or diverging : $$a_n=\cos(n/2)$$ I know that $\lim_\infty n/2= \infty $ and I also know that the cosines function is alternating between $[-1,1]$. So by a theorem I can conclude that $a_n$ is diverging.

But know I'm asking myself, if I have : $$a_n=\cos(1/n)$$ I have $\lim_\infty 1/n= 0$ then by theorem $\lim_\infty a_n = 1$, so $a_n$ is converging, right ?

but naively, I thought that sequences with cosines like both above, are always divergents because of alternating

thx

Answer 1

No, not all sequences involving sines and cosines are divergent. Here's another one:

$$a_n := \cos{2\pi n}$$

Note that the argument doesn't go to zero (it in fact tends to $\infty$), but the sequence is very convergent.

Answer 2

$\cos$ alternates as its input approaches larger and larger values in some regular way.

With the example $\cos(1/n)$, the input approaches $0$, so the alternating property just is not there.

But even more complicating, is that just because the input of $\cos$ approaches $\infty$, that is not enough to conclude you have a divergent sequence. If for example the input values, while getting larger, are approaching the same phase in the cosine cylce, then there will be a limit. Consider $\cos(1+n(2\pi+1/n^2))$. The argument gets very close to $1$ plus an integer multiple of $2\pi$, although that integer is larger with each step in $n$. But in the end, this sequence converges to $\cos(1)$.

To prove that $\cos(n/2)$ diverges is not as simple as you might like it to be. One approach is to show that $n/2$ will get very close to multiples of $2\pi$ infinitely many times, while also getting very close to multiples of $2\pi$ plus $\pi$ infinitely many times. So the sequence will have values close to $1$, and close to $-1$, infinitely many times.

Answer 3

Suppose to the contrary that the sequence $(a_n)$ converges, to $\cos\theta$. Then the sequence $(a_{n+1})$ also converges to $\cos\theta$. But since $$\cos((n+1)/2)=\cos(n/2)\cos(1/2)-\sin((n/2))\sin(1/2),$$ it follows that $\cos(\theta)=\cos(\theta+1/2)$. There is no such $\theta$.