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Math Help - Archimedean series

  1. #1
    Junior Member
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    Archimedean series

    This problem has left me scratching my head:
    The modern (and Archimedean!) meaning of “the series Σ ai (where i goes from 0 to ∞)converges to A" isusually captured by a definition like:

    (*)Σ ai (where i goes from 0 to ∞)converges to A if for every ε > 0 there is a K such that for all k K wehave (Σ ai) – A │< ε (where i goes from 0 to k).

    Archimedes himself would probably have said something more along the following lines:

    (º) Σ ai (where i goes from 0 to ∞)converges to A if both

    (1) for every L < A there is a K such that for all k K we have L < (Σ ai) (where i goes from 0 to k),
    and

    (2) for every U > A there is a Ksuch that for all k Kwe have (Σ ai) < U(where i goes from 0 to k).

    Explain, in detail, why these two definitions are actually equivalent.

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  2. #2
    Super Member
    Joined
    Apr 2009
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    Quote Originally Posted by CoraGB View Post
    This problem has left me scratching my head:
    The modern (and Archimedean!) meaning of “the series Σ ai (where i goes from 0 to ∞)converges to A" isusually captured by a definition like:

    (*)Σ ai (where i goes from 0 to ∞)converges to A if for every ε > 0 there is a K such that for all k K wehave (Σ ai) – A │< ε (where i goes from 0 to k).

    Archimedes himself would probably have said something more along the following lines:

    (º) Σ ai (where i goes from 0 to ∞)converges to A if both

    (1) for every L < A there is a K such that for all k K we have L < (Σ ai) (where i goes from 0 to k),
    and

    (2) for every U > A there is a Ksuch that for all k Kwe have (Σ ai) < U(where i goes from 0 to k).

    Explain, in detail, why these two definitions are actually equivalent.

    Attempt - Archimedes definition => Modern Definition
    for ε > 0 define L = A- ε and U = A + ε
    There is a K such that
    all k K we have L < (Σ ai)
    There is a K' such that all k K' we have (Σ ai) < U

    Let Ko = Max (K,K')
    So, for all
    k Ko we have L < (Σ ai) < U
    i.e A - ε < (Σ ai) < A + ε
    i.e.
    (Σ ai) – A │< ε

    I think other way round can be done on similar lines.
    Hence the two definitions are equivatent
    Thanks


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