Question 3.7: If 0 < a < 1 then lim a^n = 0.

We can always write a =\frac{1}{1 + b}, where b > 0. By the binomial theorem (see Exercise 2.3.14),

(1 + b)^{n} = 1+nb + \frac{n(n − 1)}{2}b^{2} + · · · + b^{n} > nb,

hence,

0 < a^{n} = \frac{1}{(1 + b)^{n}}< \frac{1}{nb}.

Since 1 / n → 0,  1 / nb → 0, and we apply Theorem 3.6 to complete the proof.