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    Every convergent sequence is Cauchy

    P-S06CA

    Area
    Prelim
    Topics
    Sequences of Numbers
    Seen at
    UGA
    Years
    2006
    Status
    draft

    1. State what it means for a sequence \(\{a_n\}\) of real numbers to be Cauchy.

    2. Prove that every convergent sequence is Cauchy.

    <1>2. Suppose \(a_n\to L\) for some \(L\in\mathbb R\), and let \(\varepsilon>0\).

    <1>3. There exists \(N\in\mathbb N\) such that \[ n\ge N\implies |a_n-L|<\frac\varepsilon2. \]

    <1>4. If \(m,n\ge N\), then \(|a_m-a_n|<\varepsilon\).

    <1>5. Therefore every convergent real sequence is Cauchy.

    Solution

    1. A sequence \((a_n)\) is Cauchy if for every \(\varepsilon>0\) there exists \(N\in\mathbb N\) such that\[ m,n\ge N\implies |a_m-a_n|<\varepsilon. \]

    This is the definition of convergence of \(a_n\) to \(L\), applied with \(\varepsilon/2\).

    By the triangle inequality and 3, \[ |a_m-a_n| \le |a_m-L|+|a_n-L| <\frac\varepsilon2+\frac\varepsilon2 =\varepsilon. \]

    The condition in 4 is exactly the Cauchy condition stated in 1.

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