Question

Use the definition of limit to prove that both of the sequences { 1/√n } and { (−1)^n/√ n } converges to 0.

Answer 1

To show that $\lim_{n \to \infty} \frac{1}{\sqrt n} = 0$

We show that given any $\epsilon > 0$ there exist $M \in \mathbb{N}$ such that $n>M \implies$ $\mid \frac{1}{\sqrt n} - 0 \mid < \epsilon$

Set $M> \frac{1}{\epsilon^{2}} \implies \epsilon > \frac{1}{\sqrt M}$

So, whenever $n> M$

Consider

$\mid \frac{1}{\sqrt n} - 0 \mid = \mid \frac {1}{\sqrt n} \mid = \frac {1}{\sqrt n} < \frac {1}{\sqrt M} < \epsilon$

as desired.

To show that $\lim_{n \to \infty} \frac{(-1)^{n}}{\sqrt n} = 0$

We show that given any $\epsilon > 0$ there exist $M \in \mathbb{N}$ such that $n>M \implies \mid \frac{(-1)^n}{\sqrt n} - 0 \mid < \epsilon$

Set $M> \frac{1}{\epsilon^{2}} \implies \epsilon > \frac{1}{\sqrt M}$

So, whenever $n> M$

Consider

$\mid \frac{(-1)^n}{\sqrt n} - 0 \mid = \mid \frac {(-1)^n}{\sqrt n} \mid \leq \mid \frac{1}{\sqrt n}\mid = \frac {1}{\sqrt n} < \frac {1}{\sqrt M} < \epsilon$

as desired.