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Math Help - Who come to beat this :)

  1. #16
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    Quote Originally Posted by RockHard View Post
    Thanks for the confirmation, so this is best shown using the property of adding two series together will only converge if both series are convergent, correct?
    That is NOT correct. For example, neither \sum_{n=1}^\infty \frac{1}{n} nor \sum_{n=1}^\infty -\frac{1}{n} converge but their sum is \sum_{n=1}^\infty (\frac{1}{n}-\frac{1}{n})= \sum_{n=1}^\infty 0 converges trivially.

    If \sum_{n=1}^\infty a_n diverges to positive infinity and b_n> 0, then \sum_{n=1}^\infty (a_n+ b_n) diverges to infinity, by the comparison test, whether \sum_{n=1}^\infty b_n converges or not.

    In fact, I would have handled the initial problem by noting that, since 1+ \frac{1}{n}> 1 \left|(1+\frac{1}{n})a_n\right|> |a_n| and so \sum_{n=1}^\infty (1+ \frac{1}{n})a_n diverges by the comparison test.
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  2. #17
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    Quote Originally Posted by HallsofIvy View Post
    since 1+ \frac{1}{n}> 1 \left|(1+\frac{1}{n})a_n\right|> |a_n| and so \sum_{n=1}^\infty (1+ \frac{1}{n})a_n diverges by the comparison test.
    I think you only proof \sum_{n=1}^{\infty}\left|\left(1+\frac{1}{n}\right  )a_{n}\right| diverges.

    I think It's not imply \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n}
    diverges
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  3. #18
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    For example:

    we take: a_{n}=\left(-1\right)^n\frac{1}{n}

    then: \sum_{n=1}^{\infty}\left|\left(1+\frac{1}{n}\right  )a_{n}\right|=\sum_{n=1}^{\infty}\left|\left(1+\fr  ac{1}{n}\right)(-1)^n\frac{1}{n}\right|\geq \sum_{n=1}^{\infty}\frac{1}{n}

    so it is diverges

    but:
    \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n}  =\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)(-1)^n\frac{1}{n}

    is a alternating series

    so obviously converges

    so \sum_{n=1}^{\infty}\left|\left(1+\frac{1}{n}\right  )a_{n}\right|

    diverges don't imply

    \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n}

    diverges
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  4. #19
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    The problem is equivalent to showing that if \sum_n \left(1+\frac{1}{n}\right)a_n converges, then \sum_n a_n converges.

    Or, that if \sum_n a_n converges, then \sum_n\frac{a_n}{1+\frac{1}{n}} converges. (Applying this result to \sum_n (1+\frac{1}{n})a_n gives the previous one)

    So, assume that \sum_n a_n converges.

    Note that we have \frac{1}{1+\frac{1}{n}}=1-\frac{1}{n}+O(\frac{1}{n^2}), hence there are n_0,C>0 (that could be explicitated, but it's no use) such that, for all n\geq n_0,

    1-\frac{1}{n}-\frac{C}{n^2}\leq\frac{1}{1+\frac{1}{n}}\leq 1-\frac{1}{n}+\frac{C}{n^2}.

    However, since \sum_n a_n converges, a_n\to_n 0 is bounded: |a_n|\leq M for some M and all n, hence \sum_n \frac{a_n}{n^2} converges absolutely.

    Moreover, it is a "classical" fact that \sum_n \frac{a_n}{n} converges (cf. below).

    As a consequence, we have

    \sum_{n=n_0}^N \frac{a_n}{1+\frac{1}{n}}-\sum_{n=n_0}^Na_n+\sum_{n=n_0}^N\frac{a_n}{n} = \sum_{n=n_0}^N C_n \frac{a_n}{n^2}

    where |C_n|\leq C. Each of the series converges, except (perhaps) the first one. Therefore, the first one converges as well.
    .

    This concludes.

    Appendix: if \sum_n a_n converges, then \sum_n\frac{a_n}{n} converges. This is a consequence of "Abel summation". Let S_n=a_1+\cdots+a_n, so that (S_n)_n converges to \sum_{n=0}^\infty a_n and is therefore bounded: |S_n|\leq M for all n and some M. We have, for all N,

    \sum_{n=1}^N \frac{a_n}{n} = \sum_{n=1}^N \frac{S_n-S_{n-1}}{n} =  \sum_{n=1}^{N-1} \left(\frac{1}{n}-\frac{1}{n+1}\right)S_n + \frac{S_N}{N}=\sum_{n=1}^{N-1} \frac{1}{n(n+1)}S_n +\frac{S_N}{N},

    and the term on the right end converges to a finite limit as N\to\infty since \frac{|S_n|}{n(n+1)}\leq \frac{M}{n^2} (general term of a convergent series), and \frac{|S_N|}{N}\leq \frac{M}{N}\to_N 0.
    Last edited by Laurent; November 28th 2009 at 01:22 AM.
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  5. #20
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    Thumbs up

    Excellent Work !!!
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