Is this proof of the divergence of a sequence correct

Question

I need to prove if $$a_n=\cos(\pi n)$$ converges or diverges. First, let's notice that $-1\leq \cos(\pi n)\leq1$, and that $\cos(\pi n)$ oscillates between $-1$ and $1$ as $n\to\infty$.

With this said, let's take the subsequence $b_n=(1, -1, 1, -1,\ldots)=(-1)^{n+1}$ of $a_n$, which also oscillates between $-1$ and $1$. If we take two subsequences of $b_n$, let them be $b'_n=1$ and $b''_n=-1$ we will find that these two subsequences do not converge to the same limit. Since $b_n$ has two subsequences that do not converge to the same limit, $b_n$ is a divergent sequence. Since $b_n$ is divergent, and by the divergence criteria for sequences, $a_n$ is then divergent.

I'm very new to sequences so probably not a very formally correct proof.

Answer 1

Yes, your proof is valid.

But

First, let's notice that $\color{red}{−1≤cos(πn)≤1}$, and that $cos(πn)$ oscillates between $−1$ and $1$

The first, in red, isn't necessarily relevant and you never used it. And it won't help you. The second, depending on the whim of the grader, may or may not need to be verified or more formally defined.

I'd say: $\cos(n\pi) = (-1)^n$ which equals $+1$ if $n$ is even, and equals $-1$ if $n$ is odd.

let's take the subsequence bn=(1,−1,1,−1,…)=(−1)n+1 of an, which also oscillates between −1 and 1.

This isn't invalid but I don't see why you are doing this. There is no reason you sequence starts on $1$ rather than $-1$. Just use $a_n$ and don't bother with this.

If we take two subsequences of bn, let them be b′n=1 and b′′n=−1

Okay, but it might be better to formally describe how to do this.

Let $b'_n = a_{2n}= \cos (2n\pi) = 1$ and let $b''_n = a_{2n+1} = \cos((2n+1)\pi) = -1$.

we will find .....

whoa... if you say "we will find" you're just asking for the grader to so "Oh, yeah. When will we find that?" :)

.... that these two subsequences do not converge to the same limit.

Just say that they do converge to different limits.

I'd be a real sadist if I required you to prove that but it's enough to say $b'_n$ converges to $1$ (because it is constant) and $b''_n$ converges to $-1$.

Since bn has two subsequences that do not converge to the same limit, bn is a divergent sequence.

You should cite the theorem that states this is so.

Since bn is divergent, and by the divergence criteria for sequences, an is then divergent.

Again there was no reason to have ever introduce the $b_n$.

Anyway... I'd give full marks but with the comments I just gave.