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     \Sigma ln(\frac{n}{n+1}) -ln(\frac{n+2}{n+3})


    The sum is from 1 to infinity.

    Usually I approach series problems by testing to see if it diverges first but after applying the L'H rule, the series does approach 0 as it goes to infinity.

    Usually from here I try the comparison/ratio/root tests. I'm guessing that the comparison test can work here in some way although I'm not able to see which series to compare it to?

    Any idea how to prove this series diverges?
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    (Original post by Zilch)
     \Sigma ln(\frac{n}{n+1}) -ln(\frac{n+2}{n+3})


    The sum is from 1 to infinity.

    Usually I approach series problems by testing to see if it diverges first but after applying the L'H rule, the series does approach 0 as it goes to infinity.

    Usually from here I try the comparison/ratio/root tests. I'm guessing that the comparison test can work here in some way although I'm not able to see which series to compare it to?

    Any idea how to prove this series diverges?
    \ln \left(\dfrac{n}{n+1}\right)-\ln \left(\dfrac{n+2}{n+3}\right)= \ln  n-\ln (n+1)+\ln (n+3)-\ln (n+2)
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    (Original post by Lord of the Flies)
    \ln \left(\dfrac{n}{n+1}\right)-\ln \left(\dfrac{n+2}{n+3}\right)= \ln  n-\ln (n+1)+\ln (n+3)-\ln (n+2)
    I got that, instead I converted both though to a single log series and took the limit from there. My lecturer tells me there's a comparison from there which can prove it diverges. I'm wondering which one it is?
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    (Original post by Zilch)
    I got that, instead I converted both though to a single log series and took the limit from there. My lecturer tells me there's a comparison from there which can prove it diverges. I'm wondering which one it is?
    The reason why I wrote it like that is because it makes it a little easier to see what's going on:

    \displaystyle\sum_1^m \ln \left(\dfrac{n}{n+1}\right)-\ln \left(\dfrac{n+2}{n+3}\right) =-\ln 3-\ln (m+1)+\ln (m+3)

    Which obviously converges to \ln \frac{1}{3}
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    (Original post by Lord of the Flies)
    The reason why I wrote it like that is because it makes it a little easier to see what's going on:

    \displaystyle\sum_1^m \ln \left(\dfrac{n}{n+1}\right)-\ln \left(\dfrac{n+2}{n+3}\right) =-\ln 3-\ln (m+1)+\ln (m+3)

    Which obviously converges to \ln \frac{1}{3}
    Thanks!.

    One last series if you could please.

     \Sigma  4^n(2+sin(n)) - \frac{4^n}{2+sin(n)}
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    (Original post by Zilch)
    Thanks!.

    One last series if you could please.

     \displaystyle\sum  4^n\left(2+\sin n - \frac{1}{2+\sin n}\right)
    Well, does the summand converge to 0?
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    (Original post by Lord of the Flies)
    Well, does the summand converge to 0?
    oscillates from 0 and infinity. Thanks, I suppose that makes it fail the divergence test.
 
 
 
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