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.
[Math] Proof that the sequence is divergent
sequences-and-series
Best Answer
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 $\epsilon\le 2$.)