Roughly the argument is as follows:
Let the general term in the sequence be bn = an
We need to show that
For every ε > 0 there exits an N such that | bN - 0| < ε i.e. |aN | <ε
(The idea is that ε is a very small number that we can make as small as we want, and that the corresponding N is a very large number)
We show this by explicitly obtaining an expression for N - this shows that it exists.
Take the natural log of both sides of |aN | <ε
N ln| a| < ln (ε)
but since a is in the open interval (-1,1), ln| a| is a negative number. When both side are divided by
ln | a|, the sense of the inequality is reversed.
N > ln(ε) / ln| a|
The right hand side is a largish positive number.
Thus the N we needed to identify is ln(ε) / ln| a| rounded up to the next integer value.