Richard P. answered • 07/17/15

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Roughly the argument is as follows:

Let the general term in the sequence be b

_{n}= a^{n}We need to show that

For every ε > 0 there exits an N such that | b

_{N}- 0| < ε i.e. |a^{N}| <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 |a

^{N}| <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.