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Math Help - series (converging)

  1. #1
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    series (converging)

    Im not sure how to do the part in red, do i use the comparison test?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    " alt="\displaystyle\sum_{r=1}^{\infty}\log\dfrac{r+1}{r} =\displaystyle\sum_{r=1}^{\infty}(\;\log({r+1})-\log {r}\" />

    (telecopic series)


    Fernando Revilla

    Edited: Sorry, I didn't read the first part. So, an alternative is:

    \log \dfrac{r+1}{r}\sim \dfrac{1}{r}\;(r\rightarrow +\infty)

    and the harmonic series is divergent, so the given series is divergent.
    Last edited by FernandoRevilla; January 19th 2011 at 04:26 AM.
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