{(2-2n)/n}
I want to use the definition of convergence to prove this converges. So I assume the sequence converges to -2. Let epsilon>0 be arbitrary. We need to find N so that when ever n>N then absolute value of(a subn-A)=
abs value((2-2n)/n+2)<epsilon
-epsilon<2/n-2n/n+2<epsilon
-epsilon<2/n<epsilon
-2/epsilon>n>2/epsilon=N
This is where I get stuck....