{a_n}\to A iff. {a_n-A}\to 0

**dwsmith** · June 18th 2011, 06:28 PM

Show that $\{a_n\}_{n=1}^{\infty}\to A$ iff. $\{a_n-A\}_{n=1}^{\infty}\to 0$

(i) $\Rightarrow$
Suppose $\{a_n\}_{n=1}^{\infty}\to A$ . Let $\epsilon>0 \ \text{and} \ \exists N\in\mathbb{N}$ such that $|a_n-A|<\epsilon$ .

$\Rightarrow |a_n-A|=|(a_n-A)-0|<\epsilon$
$\Rightarrow \{a_n-A\}_{n=1}^{\infty}\to 0$

(ii) $\Leftarrow$
Now, suppose $\{a_n-A\}\to 0$ . Let $\epsilon>0 \ \text{and} \ \exists N\in\mathbb{N}$ such that $|(a_n-A)-0|<\epsilon$ .

$\Rightarrow|(a_n-A)-0|=|a_n-A|<\epsilon$
$\Rightarrow \{a_n\}_{n-1}^{\infty}\to A$

Is this all that needs to be done?

**FernandoRevilla** · June 18th 2011, 10:16 PM

Originally Posted by dwsmith

Is this all that needs to be done?

Yes, your proof is correct.

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