∃ε ∀n0 ∀n. n > n0 ⇒ |xn - 1/2| ≥ ε
where xn = (3n-2)/(4n+1). I wrote ∀n because you did not provide a quantifier for n, and by default the quantifier is universal. In fact, the correct negation is
∃ε ∀n0 ∃n. n > n0 and |xn - 1/2| ≥ ε.
In words, there exists an ε such that there exist arbitrarily far elements of the sequence whose distance to 1/2 is at least ε. One way to prove that there exist arbitrarily far elements of the sequence with some property P is to show that all elements possess this property starting from some point. That is, to prove
∀n0 ∃n. n > n0 and P(xn) (1)
∃n0 ∀n. n > n0 ⇒ P(xn) (2)
because (2) is stronger that (1). (Why?)
To prove (2), note that xn approaches 3/4 from below. Therefore, eventually it is greater than say, the midpoint between 1/2 and 3/4, i.e., 5/8. So, take ε = 5/8 - 1/2 = 1/8 and find an n0 such that xn ≥ 5/8 for all n > n0.