Answer
No.
Counterexample: $1,3,1,3,1,3,...$
Work Step by Step
We have seen examples of sequences where terms alternate between two constant values, and we found that such sequences do not converge.
An example:
$1,3,1,3,1,3,...$
This sequence is bounded from above by 3, but is not convergent.
If we set $\epsilon=1/2$, and L is the limit of the sequence, because of the alternating 1's, it must be that $L\in(0.5,1.5),$
And because of the alternating $3$'s, $L\in(2.5,3.5).$
L can not exist because it can't be in both intervals.