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I need to prove that the sequence $(-1)^n$ for $n\in\mathbb{N}$ diverges. Is this proof correct? Is there a better way to prove it? Probably my proof is very poor, but here it is.

We prove this by contradiction. First of all, we know that $-1 \leq (-1)^n \leq 1$. Now suppose it converges to a value $a$. Then we have $$\forall \epsilon >0, \exists n_0\in\mathbb{N}: |(-1)^n-a| < \epsilon \,\,\,\, \forall n\in\mathbb{N}, n> n_0$$ in particular, take $\epsilon :=1$.

Then $\exists n_0 \in \mathbb{N}$ such that the above holds for $\epsilon =1$. Take now $n > n_0$ then we have $$|(-1)^n-a| \leq |(-1)^n|+|a| = |-1|^n+|a|=1+|a| \geq 1$$ where I have used the triangle inquality and the fact that $|b^n| = |b|^n$ and that $|a| \geq 0$.

Hence we have a contradiction because this doesn't hold for any $\epsilon >0$.

Is this proof correct? Because in my book the solution is quite different. It involves diving into cases where $n$ is odd or even. I am pretty sure I can't do the triangle inequality part, but I have no other idea since I don't understand the proof in the book.

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    I don't understand why you think you've reached a contradiction. You've obtained $|(-1)^n - a| \leq 1 + |a|$ and $1 + |a| \leq 1$, but so what? – 2017-01-22
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    no, I've obtained $1+|a| \geq 1$ but by the definition I should have $1+|a| < 1$ – 2017-01-22
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    my problem is that I've used the triangle inequality and that step is certainly correct. However, it is not correct to say that there is a contradiction after that. Indeed I could have $|(-1)^n-a| < \epsilon \leq |(-1)^n| + |a|$ – 2017-01-22
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    well, actually murray is onto something, you have proved $$ |(-1)^n-a| \leq |(-1)^n|+|a| =1+|a|$$ and nothing more. No contradiction so far. You better choose look at $n$ and $n+1$, how does the neighborhood of $a$ changes? – 2017-01-22
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    Your solution is not right because you found something like $p\le c$ and $c \ge 1$. There is no contradiction. – 2017-01-22
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    @Arnaldo yeah you are correct I realised it. I got confused somehow. I am trying to figure out if the proofs down are rigorous enough now – 2017-01-22
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    If you have seen Cauchy sequences yet, it is also very easy to show that this sequence is not Cauchy and therefore does not converge. – 2017-01-22
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    @Euler_Salter: no problem. If I need clarify anything, please let me know. – 2017-01-22
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    @Vik true, would it work with any $\epsilon \leq 2$ right? Because then I just take any two $n$ with a step of $2$, like $2n_0$ and $2n_0+2$ and I get a contradiction? – 2017-01-22
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    You have the right idea, but I think you mean a step of 1, rather than a step of 2. – 2017-01-22
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    well if they are either both even or both odd (which occurs at a step of two) then you either have $2$ or $-2$ which then would give |\pm 2| = |2| \leq 2$ so a contradiction. Otherwise, what did you have in mind? – 2017-01-22
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    No, if you take two elements of the sequence that are either both at even indices or both at odd indices, then those elements are equal. You need to look at one element from an even index and one from an odd index. – 2017-01-22

3 Answers 3

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If ${\{a_n\}}_{n = 1}^{\infty}$ converges to $l$, where $a_n = {(- 1)}^n$ for all $n = 1 , 2 , \ldots$, all the subsequences of ${\{a_n\}}_{n = 1}^{\infty}$ converge also to $l$. In particular ${\{a_{2 n + 1}\}}_{n = 1}^{\infty}$, given by $a_{2 n + 1} = {(- 1)}^{2 n + 1} = - 1$ for all $n = 1 , 2 , \ldots$, converges to $l = - 1$, but ${\{a_{2 n}\}}_{n = 1}^{\infty}$, given by $a_{2 n} = {(- 1)}^{2 n} = 1$ for all $n = 1 , 2 , \ldots$, should converge also to $l$ and it's false clearly because $\lim_{n \to \infty} a_{2 n} = \lim_{n \to \infty} 1 = 1 \neq - 1 = l$. This argument proves that ${\{a_n\}}_{n = 1}^{\infty}$ doesn't converge.

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    Yeah that was what I was looking for! Just to clarify , is this method of the subsequences the standard method for showing a sequence is divergent, or is it the "standard" method for alternating sequences like this one? – 2017-01-22
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    I think it can serve both types but always depends on each case. – 2017-01-22
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Another way to see that is

$$a_{n+1}=-a_n \to a_{n+1}+a_n=0$$

And if $a_n$ converge to L then we should have

$$L+L=0\to L=0$$

What is not true because $a_n=\pm 1$.

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    I've never seen this way of proving that sequence converge/diverge. Is there a way to write it more rigorously and formally? Should I talk about subsequences? – 2017-01-22
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    It is formally written. I just wrote the sequence by recurrence. That is a good way to try to find the limit when it exist. – 2017-01-22
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    @Arnaldo you just need to state that $f:x \mapsto -x$ is continuous, so $$a_n \to a \implies f(a_n) \to f(a)$$ – 2017-01-22
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    @Baconaro , in this case, what would the function be? – 2017-01-22
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    @Baconaro: I don't need that, because it is equivalen to $a_{n+1}+a_n=0$ – 2017-01-22
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    @Euler_Salter: it is not necessary. Take another look in my answer. – 2017-01-22
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    @Arnaldo yeah, sorry, cant read. – 2017-01-22
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One way: suposse $a_n\to L\in\mathbb{R}.$ Taking $\epsilon =1,$ there exists $n_0$ natural number such that if $n\ge n_0$ then, $\left | (-1)^n-L\right |<1.$ For $n=2n_0$ we get $\left | 1-L\right |<1,$ so, $L>0.$

For $n=2n_0+1,$ we get $\left | -1-L\right |=\left | 1+L\right |<1,$ which implies $L<0$ (contradiction).

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    this is very clever! I like it – 2017-01-22
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    otherwise, could I sum them up? Finding $2 = |1+1| = |(1+L+1-L| \leq !1-L|+|1+L| < 2$ so this is a contradiction? – 2017-01-22
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    Right, that is another way. – 2017-01-22