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Math Help - Who come to beat this :)

  1. #1
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    Thumbs up Who come to beat this :)

    If \sum_{n=1}^{\infty}a_{n} diverges

    show that:

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

    diverges
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  2. #2
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    I never get these general term series question, multiplied by some function
    I know that simply by the divergence test or that it is the harmonic series.

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

    \lim_{n\to\infty} 1 + lim_{n\to\infty} \frac{1}{n} =1

    which is not 0 and by the divergence test if the limit of a_n \ne 0 diverges

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

    Could say that the An is just like multiplying a constant and which multiplying by a constant will not affect if a series diverges or convergences? just the sum?

    \sum_{n=1}^{\infty}(1+\frac{1}{n})a_n

    similarly

    \sum_{n=1}^{\infty}\frac{a_n(1+n)}{n}

    This is a Harmonic series which diverges...I am incorrect?
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  3. #3
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    \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n} = \sum_{n=1}^{\infty} a_{n} + \sum_{n=1}^{\infty}\left(\frac{1}{n}\right)a_{n}.

    What you have here is either a sum of two divergent series or a sum of a convergent and a divergent series, both of which diverge.
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  4. #4
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    Defunkt, would the way I TRIED, lol, to show this be applicable? more specifically the last bit of my reply
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  5. #5
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    Well, I don't really understand what you're saying there :P Try organizing your thoughts more clearly and then presenting them.

    If you are saying that multiplying the series by \left( 1 + \frac{1}{n} \right) is like multiplying by a constant, that's wrong. It simply is not a constant. And no, it is not a harmonic series -- http://en.wikipedia.org/wiki/Harmoni...s_(mathematics) this is the only form of a harmonic series.
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  6. #6
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    Thanks for the tips, I am just so eager to learn, I make a mess of myself, I was trying to say that multiplying by A_n to the series of
    \sum_{n=1}^{\infty}(1+\frac{1}{n}), you can treat A_n as constant, more than likely this is where I am incorrect, and that multiplying a constant to a series will not affect its convergence or divergence just it's sum.

    Also tried saying that

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

    can be written as

    \sum_{n=1}^{\infty}\frac{n+1}{n}

    which looks to be of similar form to the harmonic series to me, but what the hell, I am more incorrect than correct, when I post messes like these, lol.
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  7. #7
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    The harmonic series is ONLY this: \sum_{n=1}^{\infty}\frac{1}{n}. NOTHING else. Not everything that is divided by n is a harmonic series :P

    The part about multiplying by a constant is true, though. If you multiply a series by a non-zero constant, its convergence will not change -- only the value of the series (if it converges). The thing here is that 1+ \frac{1}{n} is not a constant (or do you mean the original divergent series \sum_{n=1}^{\infty}a_n? either way, both aren't constants), even though it goes to 1 as n goes to infinity.
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  8. #8
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    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?
    Last edited by RockHard; November 26th 2009 at 10:34 PM.
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  9. #9
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    Quote Originally Posted by Defunkt View Post
    \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n} = \sum_{n=1}^{\infty} a_{n} + \sum_{n=1}^{\infty}\left(\frac{1}{n}\right)a_{n}.

    What you have here is either a sum of two divergent series or a sum of a convergent and a divergent series, both of which diverge.
    If \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n} converges

    we can decompose it into two part:

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

    In other words, we can treat it as finite summation.

    but if it diverges, I think we can't treat it as finite summation, so If

    \sum_{n=1}^{\infty}a_{n} diverges,

    can we have:
    \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n}  =\sum_{n=1}^{\infty}a_{n}+\sum_{n=1}^{\infty}\frac  {a_{n}}{n}
    ?
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  10. #10
    Senior Member Shanks's Avatar
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    let a_n^{+}=max\{a_n,0\}, a_n^{-}=max\{-a_n,0\}
    then \sum_{n=1}^{\infty} a_{n}=\sum_{n=1}^{\infty} a_{n}^{+}-\sum_{n=1}^{\infty} a_{n}^{-}.
    Since \sum_{n=1}^{\infty} a_{n} divergens, then at least one of them diverges.
    If only one of them diverges; then obviously the conclusion is true.
    If both of them diverges, thennnnnnn......
    what happens if \sum_{n=1}^{\infty} a_{n} diverges to +\infty(\text{or}-\infty)?
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  11. #11
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    Well, I think that if the way Defunkt looked at the function \sum_{n=1}^{\infty}(1+\frac{1}{n})a_n as a sum of two series, and the property is for that too be convergent both series he broke apart must be convergent
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  12. #12
    Senior Member Shanks's Avatar
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    maybe, we can think in a inverse direction, maybe it is easier to prove:
    If \sum_{n=1}^{\infty} (1+\frac{1}{n})a_{n} converges,
    then \sum_{n=1}^{\infty} a_{n} converges.
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  13. #13
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    Err.. yes, my mistake. It would probably be easier to prove noting that
    1) \sum_{n=1}^{\infty} a_{n} diverges
    2) \left( 1 + \frac{1}{n} \right)a_{n} \geq a_n \ \forall n \in \mathbb{N}
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  14. #14
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    Talking

    Quote Originally Posted by Defunkt View Post
    Err.. yes, my mistake. It would probably be easier to prove noting that
    1) \sum_{n=1}^{\infty} a_{n} diverges
    2) \left( 1 + \frac{1}{n} \right)a_{n} \geq a_n \ \forall n \in \mathbb{N}
    If a_{n}>0 for \forall n \in \mathbb{N}^{+}

    then we have:
    \left( 1 + \frac{1}{n} \right)a_{n} \geq a_n \ \forall n \in \mathbb{N}

    so......I think "2)" will not correct for all n
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  15. #15
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    lol, sorry again. shouldn't post so early in the morning. the equality will hold in absolute value, and that should work... off i go now :S
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