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

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

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

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

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

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

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

Is this all that needs to be done?

2. Re: {a_n}\to A iff. {a_n-A}\to 0

Originally Posted by dwsmith
Is this all that needs to be done?