# Is this proof of the divergence of a sequence correct?

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.

• The two subsequences you took, $\;b_n,\,b'_n\;$ give the exact same values as $\;\cos x\;$ is a even function...Thus your proof doesn't yield what you wrote. Jan 11, 2020 at 22:50
• Sorry if I didn't follow, but what do you mean when you say the "give" the exact same values as $cos x$? I thought all that mattered for this subsequences was what was their limit as $x \to\infty$, and the fact that these limits differ. Jan 11, 2020 at 23:01
• This would be valid IF you can explain why a sequence having two subsequences that converge that do not converge to the same limit means the sequence diverges. Such an explaination can be "According to Theorem 17.somethingerother: If a sequence converges to a limit, then all its subsequences will also converge to that limit. As that is not the case, this sequence does not converge". BTW, not every text defines "diverge" and "doesn't converge", but many, maybe most, do. Does yours? Jan 11, 2020 at 23:23
• Actually, I'm wrong about the definition of divergence. Divergent does mean "not convergent". Always. Not sure why I have a mental block against that. Jan 11, 2020 at 23:55
• I did say "by the divergence criteria for sequences", which would be the equivalent of y our "according to theorem 17.something". At least in my text book the theorem is simply called "Divergence criteria for sequences". Jan 12, 2020 at 0:23

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.

• Thanks, I do see now how flawed my demonstration was. I will do it again and hopefully better. There were things I did not see; for example, I did the distintion between $a_n$ and $b_n$ because I confused the domain of $a_n$ to be $\mathbb{R}$ (!), so I did not notice they are the same! That is how lost I was haha. Quite embarassing actually. Thanks for all the notes, fleablood! Jan 12, 2020 at 0:29
• I wouldn't say it was flawed. And I was sincere when I said I'd give you full marks. I just think it could be neatened up is all. And "neatness" comes with practice and experience. Jan 12, 2020 at 1:07
• Thanks fleablood! :) Jan 12, 2020 at 2:38

Hint:

$$\text{For}\;\;n\in\Bbb Z\;,\;\;\cos\pi n=(-1)^n$$

• This is a great hint, thanks Antonio. Jan 11, 2020 at 23:01

You need to remember basic trigonometry for this one: $$\begin{array}{cc} n & \cos(n\pi) \\ \hline 1 & -1 \\ 2 & +1 \\ 3 & -1 \\ 4 & +1 \\ \vdots & \phantom{-}\vdots \end{array}$$