Question

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

Answer 1

Solution 1:

We know that $\{a_n\}$ converges to L, or, $\lim_{n \rightarrow \infty}=L$ if $\forall$ $\epsilon$ $>0$ $\exist$ N$\in$ $Z^+$ such that $\forall$ $n>N$,

$|a_n-L|<\epsilon$ .

Consider $|\dfrac{1}{\sqrt n}-0|=|\dfrac{1}{\sqrt n}|=\dfrac{1}{\sqrt n}<\epsilon$

$\Rightarrow 1<\epsilon\cdot\sqrt n \Rightarrow \dfrac{1}{\epsilon}<\sqrt n \Rightarrow n>\dfrac{1}{\epsilon^2}$

Proof: Let $\epsilon$ $>0$ . Choose N$>\dfrac{1}{\epsilon^2}$ . Then, $\forall$ $n>N$, $|\dfrac{1}{\sqrt n}-0|=|\dfrac{1}{\sqrt n}|$

Also, $n>N>\dfrac{1}{\epsilon^2} \Rightarrow \epsilon^2>\dfrac1 n \Rightarrow \sqrt{\dfrac1n}<\epsilon \Rightarrow {\dfrac1{\sqrt n}}<\epsilon$

Thus, $|\dfrac{1}{\sqrt n}-0|=\dfrac{1}{\sqrt n}<\epsilon$

Hence, by above statement, $\{\dfrac{1}{\sqrt n}\}$ converges to 0.

Solution 2:

Consider $|\dfrac{(-1)^n}{\sqrt n}-0|=|\dfrac{(-1)^n}{\sqrt n}|=\dfrac{1}{\sqrt n}<\epsilon$

$\Rightarrow 1<\epsilon\cdot\sqrt n \Rightarrow \dfrac{1}{\epsilon}<\sqrt n \Rightarrow n>\dfrac{1}{\epsilon^2}$

Proof: Let $\epsilon$ $>0$ . Choose N$>\dfrac{1}{\epsilon^2}$ . Then, $\forall$ $n>N$, $|\dfrac{(-1)^n}{\sqrt n}-0|=|\dfrac{(-1)^n}{\sqrt n}|=\dfrac{1}{\sqrt n}$

Also, $n>N>\dfrac{1}{\epsilon^2} \Rightarrow \epsilon^2>\dfrac1 n \Rightarrow \sqrt{\dfrac1n}<\epsilon \Rightarrow {\dfrac1{\sqrt n}}<\epsilon$

Thus, $|\dfrac{(-1)^n}{\sqrt n}-0|=|\dfrac{(-1)^n}{\sqrt n}|=\dfrac{1}{\sqrt n}<\epsilon$

Hence, by above statement, $\{\dfrac{(-1)^n}{\sqrt n}\}$ converges to 0.