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Prove that the sequence $ a_n = q^n , q \in \mathbb{R} $ is convergent or divergent .
I need some explanations for some cases:
a)If $q = -1$ (is a divergent sequence).
Suppose $a_n = (-1)^n $ is convergent.
With the Cauchy criterion:
$\forall \epsilon > 0, \exists N \in \mathbb{N},\forall n,m \geq N: $
$|a_n-a_m| < \epsilon $
Then: Choose $\epsilon = 1$ (after the solution =afs)
Now my question is why shall I choose 1 for epsilon?
And if I have choosen 1:
$\forall \epsilon > 0, \exists N \in \mathbb{N},\forall n,m \geq N: $
$|a_n-a_m| < 1 $
Choose m=N+1 then $n,m \geq N$;
So I have: $ |(-1)^n-(-1)^m|=2 $ (ats)
=> Contradiction $2 \geq 1$
Why is m=N+1 choosen? and finally why do I only look at the case: $|(-1)^n-(-1)^m|$ is 2,
shouldn't I also write the case for n,m = even?
Thanks for explanations.

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up vote 2 down vote accepted

The choice $\epsilon=1$ is partly arbitrary: any value satisfying $0<\epsilon\le 2$ will work in this proof. $1$ is a simple, convenient number in that interval.

Similarly, the choice of $m=N+1$ is partly arbitrary: any $m\ge N+1$ will work. However, you’ve stated this part of the argument incorrectly. After setting $m=N+1$, the argument should continue by observing that $|a_N-a_m|=\left|(-1)^N-(-1)^{N+1}\right|=2\ge 1=\epsilon$. The point is to show that for this particular choice of $\epsilon$, namely, $\epsilon=1$, it is not true that $|a_n-a_m|<\epsilon$ for all $m,n\ge N$; specifically, it’s not true for $n=N$ and $m=N+1$. And this is the case no matter how big we make $N$, so this $\epsilon$ doesn’t ‘work’: it doesn’t have the property that every positive $\epsilon$ must have in order for the sequence to converge.

We don’t look at indices $m,n\ge N$ for which $|a_n-a_m|$ is less than $\epsilon$ because they don’t matter: we’re trying to show that $\epsilon=1$ is a ‘bad’ $\epsilon$, meaning that no matter how big we choose $N$, we can find at least one pair of $m,n\ge N$ such that $|a_n-a_m|\ge\epsilon$. And we’ve done that: $n=N$ and $m=N+1$ do the job. (Note in connection with your first question that they would do the job for any choice of positive $\epsilon\le 2$.)

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Ok, and the interval for epsilon is ]0;2] because 2 is the supremum of {|a_N-a_m|}?And that means if I have an other sequence where I have to show the same, I always take a number x out of the "epsilon-interval" and then m=N+x, to prove convergence? – phil Dec 10 '12 at 0:30
@phil: Better: because no matter what $N$ is, there is an $m\ge N$ such that $|a_N-a_m|=2$. If the supremum were not attained, you’d have to limit $\epsilon$ to the interval $(0,2)$. If a sequence diverges, your plan should be to find an $\epsilon>0$ such that there are differences $|a_m-a_n|\ge\epsilon$ arbitrarily far out in the sequence, i.e., with arbitrarily large $m$ and $n$. You won’t always be able to choose $m$ and $n$ as $N$ and $N+k$ for some $k$. – Brian M. Scott Dec 10 '12 at 0:36

I'll restate your argument here in the hope of making it clearer.

In the case of the sequence $(-1)^n$, the values bounce between $-1$ and $1$. The Cauchy criterion requires that if you go far enough out, the values all get very close together. Clearly they don't do that, because any two consecutive values are $2$ apart.

To make this argument more rigorous, if the sequence is Cauchy, then choosing $\epsilon = 1$, there must be an $N$ so that all $a_i$ after $N$ are at most $1$ apart. But for any value of $N$, $a_{N+1}$ is $2$ apart from $a_N$. So there is no such $N$, and therefore the sequence is not Cauchy.

So, why do we choose $\epsilon = 1$? Just because it's a number less than $2$ that's easy to work with. We could have chosen $1.9$ or $0.001$ and the argument would work just as well.

Why do we choose $m = N + 1$? Just because it's an example of a value that makes $a_m$ not too close to $a_N$. Other values that would also work are $m = N + 3$ or $m = N + 5$. We just have to find one example that makes the Cauchy criterion fail.

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