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Thread: Nondecreasing Partial Sums

  1. #1
    Member RedBarchetta's Avatar
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    Nondecreasing Partial Sums

    Suppose that...

    $\displaystyle
    \sum\limits_{n = 1}^\infty {a_n }
    $ is an infinite series with $\displaystyle
    a_n \geqslant 0
    $ for all n. Then each partial sum is greater than or equal to its predecessor because $\displaystyle
    s_{n + 1} = s_n + a_n
    $

    $\displaystyle
    s_1 \leqslant s_2 \leqslant s_3 \leqslant \cdot \cdot \cdot \leqslant s_n \leqslant s_{n + 1} \leqslant \cdot \cdot \cdot
    $

    Would anyone mind explaining to me how $\displaystyle
    s_{n + 1} = s_n + a_n
    $ is true?

    Let's say we have: $\displaystyle
    \sum\limits_{n = 1}^\infty {n^2 }
    $, how about letting n=2.

    Let's write out the first couple terms:

    $\displaystyle
    \sum\limits_{n = 1}^\infty {n^2 } = 1 + 4 + 9 + \cdot \cdot \cdot + n
    $

    So....

    $\displaystyle
    \begin{gathered}
    s_{n + 1} = s_n + a_n \hfill \\
    s_3 = s_2 + a_2 \hfill \\
    14 \ne 5 + 4 \hfill \\
    \end{gathered}
    $

    What am I not getting?

    Thank you.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    Suppose that...

    $\displaystyle
    \sum\limits_{n = 1}^\infty {a_n }
    $ is an infinite series with $\displaystyle
    a_n \geqslant 0
    $ for all n. Then each partial sum is greater than or equal to its predecessor because $\displaystyle
    s_{n + 1} = s_n + a_n
    $
    That's not correct. It should be $\displaystyle s_{n + 1} = s_n + a_{n+1} $ ($\displaystyle s_n$ is the sum of the first n terms; you need to add the (n+1)th term to that in order to get the sum of the first (n+1) terms).
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  3. #3
    Member RedBarchetta's Avatar
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    Thank you. Just a misprint in the book.
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