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We just had our first calculus lecture, and I'm kinda stuck at this proof right now:

Prove that $\lim_{n\to\infty}\{(-1)^n\}$ diverges.

Given: $\lim_{n\to\infty}(-1)^n = a$. Thus for $\varepsilon_0 = 1 $ there should be $n_0 \in\mathbb N$ so that $$\forall n > n_0; |(-1)^n-a| < 1 $$ But when $n > n_0$, then $$|(-1)^n-(-1)^{n+1}| \le |(-1)^n-a|+|a-(-1)^{n+1}| < 1+1 = 2$$ -> A contradiction!

Now I think I lose it at $$|(-1)^n-(-1)^{n+1}|$$ Why does he substitute 'a' with $(-1)^{n+1}$, or does he anyways?

any help is appreciated !

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4  
If you've quoted the question correctly, it's badly formulated. It's the sequence that diverges, not the limit. It should either say "$\{(-1)^n\}_n$ diverges" or "$\lim_{n\to\infty}(-1)^n$ doesn't exist" (no need for the curly braces in that case). –  joriki Sep 12 '12 at 11:13
    
Agree with joriki's comment. –  Seyhmus Güngören Sep 12 '12 at 11:14
    
Here is a fun problem. If $a_n$ is a convergent sequence, then the sequence $|a_{n+1}-a_n|$ converges to zero. –  Baby Dragon Sep 12 '12 at 11:22

4 Answers 4

up vote 1 down vote accepted

This is not a substitution; it's an application of the triangle inequality $|a-b|\le|a-c|+|c-b|$.

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i see how that is in there.. i don't see, however, the relationship to triangles here (?) and why he starts of with |(-1)^n - (-1)n+1| to begin with.. :\ –  foaly Sep 12 '12 at 11:40
    
@compul: The term triangle inequality is generalized from the triangle inequality in the Euclidean plane, which states that any two side lengths of a Euclidean triangle must add up at least to the third side length. It holds in any metric space, and in particular in $\mathbb R$. You can regard the segments $ab$, $ac$ and $cb$ as forming a degenerate triangle with colinear vertices. The reason the line begins with $|(-1)^n-(-1)^{n+1}|$ is that this is known to be $2$. –  joriki Sep 12 '12 at 11:49
    
i think i just got that :D –  foaly Sep 12 '12 at 12:04

You know that there is a theorem stating:

If $\{a_n\}$ converges then every subsequence of it converges.

Therefore, if you can find two subsequences of a sequence $\{a_n\}$ which converge to the different limits, then the sequence $\{a_n\}$ does not converge! Here are two converging subsequences $1, 1, 1, 1, ...$ and $-1, -1, -1, -1, ...$ which converge to $1$ and $-1$ respectively. And, $1\neq -1$.

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should it be $an=a_n$ ? :) –  clark Sep 12 '12 at 11:21
    
@clark: Thanks friend for noting me that. :) –  Babak S. Sep 12 '12 at 11:23
    
Your proof uses a stronger theorem than you quote, namely that if $\{a_n\}$ converges then each of its subsequences converges to the same limit. –  joriki Sep 12 '12 at 11:26
    
@joriki: Yes exactly. I thought the OP would give his answer by a part of the theorem; however, he really wanted to know about the approach based on your answer instead. Thanks. –  Babak S. Sep 12 '12 at 11:37
    
I like this; another "must know" theorem! $\quad\ddot\smile\quad$ –  amWhy Mar 17 '13 at 0:49

Recall the definition of convergence. For every $\epsilon$ their is an $n_0$ such that FOR ALL $n>n_0$, $|a_n-a|<\epsilon$. Now Pick $\epsilon=1$ (what you called $\epsilon_0$). Then for some $n_0$, $|a_n-a|<1$ (since 1 is our $\epsilon$). Thus $n+1$ is greater than $n$. This is why you instructor may make that substitution. $$1=|(-1)^n-(-1)^{n+1}| \leq |(-1)^n-a|+|a-(-1)^{n+1}| < 1+1 = 2$$ Note that the first inequality is just the triangle inequality. Now for the second. If $n>n_0$ ($n_0$ being the n-value that makes $\epsilon=1$), then $n+1>n>n_0$. Since both numbers are greater than $n_0$, we get the two inequalities, $$|(-1)^n-a|,|a-(-1)^{n+1}|<1 $$.

At the risk of being redundant let me again say why your instructor may use $n+1$. It is because if $n_0$ is the n-value that make $\epsilon=1$, and $n>n_0$, then $|a_n-a|<\epsilon = 1$. But as $n+1$ is also greater than $n_0$, then $|a_{n+1}-a|<\epsilon =1 $ , as well.

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ok so i think i'm missing a lot of basics. (I'm studying on a different continent than my home continent, and didn't see any math for the last two years). * 'n0 is the n-value that make ϵ=1' - what do you mean by 'that makes ϵ=1'? * I see that n+1 is greater than n, but I don't quite see why he would just go ahead and turn the a to (−1)^n+1. @joriki said something about the triangle inequality, which i don't know much about. * to be short: i guess I'm not quite sure what this all is about. The prof basically rushed in, wrote things on the board, and rushed out again. –  foaly Sep 12 '12 at 11:55
    
Let us look at the definition of convergence. Defn: For every $\epsilon >0$ their exists $n_0$ such that for all $n>n_0$, $|a_n-a|<\epsilon$. So let us start withe the first piece: It states that For every $\epsilon$ their is an $n_0$ such that STUFF HAPPENS (we will suppress that for now). So what I mean is that if you have some $\epsilon$ (in your case $\epsilon =1$) then their is some $n_0$. So what I meant by that phrase is that $n_0$ will be dependent upon the chosen $\epsilon$. –  Baby Dragon Sep 12 '12 at 15:40

By the definition of limit, calling $p_n=(-1)^n$, you know that $p_n$ is as close as you want to its limit, definitely. When $n$ is even, $p_n=1$, and it must be as close as you want to a fixed number. This number must be $1$. By the same token, when $n$ is odd, $p_n=-1$ must be as close as you want to a fixed number, and therefore this number must be $-1$. Since limits are unique, you conclude that no limit can exist for $\{p_n\}_n$.

It is just a reformulation of the "subsequence principle" cited by @BabakSorouh in his answer, but in this particular case it can be shown by hands.

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