Which negation of the definition of a null sequence is correct? A Cauchy sequence $a_n$ is said to be a null sequence if for every $\varepsilon>0,$ there exists an integer $N$ s.t. $\forall n >N, \lvert a_n \rvert < \varepsilon$.
I thought the negation of this statement would be "There exists $\varepsilon >0$  s.t. for all integer $N$ there exists $n >N,$ $\lvert a_n \rvert \geq \varepsilon$". But I just saw the following statement:
"There exists $\varepsilon >0$ and there exists $N$ s.t. for all $n>N, \ \lvert a_n \rvert \geq \varepsilon$."
Are these two statements equivalent? I thought the second statement is more stringent than the first. If they are not equivalent, which one is correct? Thanks in advance.
 A: A null sequence by your definition is essentially a sequence whose limit is zero.  For general sequences the negation of this would be a sequence who either has no limit or whose limit is nonzero.  The sequence $a_n = (1,0,1,0,1,0,...,n\mod 2,...)$ would certainly not fall under the definition of a null sequence and as such should fall under the definition of the negation since for $\epsilon = \frac{1}{2}$, whichever $N$ you choose, you would have either $a_{N+1}$ or $a_{N+2} = 1 > \frac{1}{2} = \epsilon$.
You can check that this works for the first negation using the same logic outlined above.
In comparison, this general sequence would not fit the description of the second of the potential negations, since it is not true that for all $n>N$, $|a_n|\geq \epsilon$ since every other term is zero and would necessarily be less than $\epsilon$.
What would make the second negation work is if we have an added condition that $a_n$ converges to a limit.  In the case that $a_n$ converges to a non-zero limit, you can in fact find such $\epsilon$ and $N$ to fit the statement.
-edit- as pointed out since the statement requires the sequence be cauchy, i.e. converges to a limit, the added condition is automatically present and so the two statements are in fact equivalent.  My mistake on missing that word at the beginning.  Thanks for correcting me.
