Real analysis: Suppose that the sequence a(subn) converges?

Question

to an and that a > 0. Show that there is an index N such that a(subn) >0 for all indices n>= N

Answer 1

let´s write epsilon = &
so for n>N which depends on &
a-&< a_n < a+& .
As we can choose & arbitrarily we can take so a-&>0 and so find N
so a_n from This N is >0

Answer 2

Let {a_n} be our sequence, and a be the point to which the sequence converges. Now choose ε = a/4.

From the definition of convergence, we know there exists an integer N such that for all n>=N,
|a_n - a|<ε

i.e.,
a_n is within a/4 of a; and since a is positive, all such a_n are >0.