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Math Help - Monotonically Decreasing Sequence

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    Junior Member Kanwar245's Avatar
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    Monotonically Decreasing Sequence

    Suppose (an) is a monotonically decreasing sequence of positive numbers, and that \Sigma\limits_{n=1}^\infty a_{n} converges.

    Show that \lim_{n\rightarrow\infty} na_{n} = 0
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    Re: Monotonically Decreasing Sequence

    This is a well known theorem in the series of positive terms.
    If the series Σan of positive monotone decreasing terms is to converge then we must have not only lim(an)=0 but also lim(n(an))=0 . However the condition lim(n(an))=0 is only necessary , not a sufficient one for these type of series. If lim(nan) does not tend to zero then definitely the series diverges but lim(nan)=0 does not necessarily implies anything as to the possible convergence of the series. in fact the Abel series Σ(1/nlogn) diverges though lim(nan)=0

    Get a good book on series to revise all these theorems.

    MINOAS
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    Junior Member Kanwar245's Avatar
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    Re: Monotonically Decreasing Sequence

    Quote Originally Posted by MINOANMAN View Post
    This is a well known theorem in the series of positive terms.
    If the series Σan of positive monotone decreasing terms is to converge then we must have not only lim(an)=0 but also lim(n(an))=0 . However the condition lim(n(an))=0 is only necessary , not a sufficient one for these type of series. If lim(nan) does not tend to zero then definitely the series diverges but lim(nan)=0 does not necessarily implies anything as to the possible convergence of the series. in fact the Abel series Σ(1/nlogn) diverges though lim(nan)=0

    Get a good book on series to revise all these theorems.

    MINOAS
    Can't I use the fact that since \Sigma\limits_{n=1}^\infty a_{n} converges implies \lim_{n\rightarrow\infty}a_{n} = 0

    So, \lim_{n\rightarrow\infty}na_{n} = \lim_{n\rightarrow\infty}n\lim_{n\rightarrow\infty  }a_{n} = \lim_{n\rightarrow\infty}n * 0 = 0
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    Re: Monotonically Decreasing Sequence

    No, that definitely doesn't work.

    But you know (by definition) that \sum_{n=1}^\infty{a_n} = \lim_{n\to\infty}\sum_{k=1}^n{a_k}. So you just need to show that the finite sum is less than na_n.

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