Virtual Math Tutor: Math problem #221 (Solution)

## Wednesday, November 2, 2011

### Math problem #221 (Solution)

Solution to Math problem #221:

In view of the definition of limit of a sequence, in order to prove that $\lim_{n\to\infty}q^n=0$, we have to show that for any $\epsilon > 0$ there exists a natural number $N$ such that for any natural number $n > N$ the inequality holds true

$$|q^n - 0| <\epsilon.$$ In the case when $q = 0$ the inequality is obviously true. Let $q \not=0$. Since $0<|q|<1$, we have $1/|q| >1$. Therefore there exists a positive number $\alpha$ such that $1/|q| = 1+ \alpha$. Since $\alpha >0$, it follows from the Bernoulli's inequality (see Math problem #35) that

$$\frac{1}{|q|^n} = \left(\frac{1}{|q|}\right)^n = (1+\alpha)^n \ge 1+n\alpha > n\alpha.$$

Hence, we get

$|q|^n < \frac{1}{n\alpha}$ for any natural $n \in \mathbb{N}$. Now choose a natural number $N$ so that $$N>\frac{1}{\alpha\epsilon},$$

where $\alpha = \frac{1}{|q|} - 1$. Then for any $n > N$ we have

$$n > \frac{1}{\alpha\epsilon},$$

or,

$$\frac{1}{\alpha n} < \epsilon,$$ which yields $$|q^n-0| = |q^n| = |q|^n < \frac{1}{n\alpha} < \epsilon.$$ Finally, we have that for any positive number $\epsilon$ (however small) there exists a natural number $N$ such that for any natural $n > N$ the inequality holds true

$$|q^n - 0|<\epsilon,$$

which means exactly that $\lim_{n\to\infty}q^n = 0.$ The problem is now completely solved.