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

  1. #1
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    {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?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: {a_n}\to A iff. {a_n-A}\to 0

    Quote Originally Posted by dwsmith View Post
    Is this all that needs to be done?
    Yes, your proof is correct.
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