For x!=0 let:
a_n = sqrt(4n^2x^2-1)/(4n^2)
and:
b_n = 1/(2nabsx)
Now:
lim_(n->oo) a_n/b_n = lim_(n->oo) 2nabsx sqrt(4n^2x^2-1)/(4n^2)
lim_(n->oo) a_n/b_n = x lim_(n->oo) sqrt(4n^2x^2-1)/(2n)
lim_(n->oo) a_n/b_n = absx lim_(n->oo) sqrt((4n^2x^2-1)/(4n^2))
lim_(n->oo) a_n/b_n = absx lim_(n->oo) sqrt(x^2-1/(4n^2))
lim_(n->oo) a_n/b_n = x^2
Based on the limit comparison test, for any x!=0 the limit is finite and non null, so that the series must have the same characteristic.
But:
sum_(n=0)^oo 1/(2nabsx) = 1/(2absx)sum_(n=0)^oo1/n
is divergent, so also:
sum_(n=0)^oo sqrt(4n^2x^2-1)/(4n^2)
is divergent.