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Math Help - Infinite Series: Divergent or Convergent?

  1. #1
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    Infinite Series: Divergent or Convergent?

    Infinite
    Σ ln(n/(n+1))
    n=1

    I know that you have to test and make sure that "a sub n" goes to 0 as n goes to infinite. You could split the summation into Σln(n) - Σln(n+1) and then take the limits as n approaches infinite of each "a sub n" separately, right? You get infinite minus infinite, though. But isn't it possible that these two limits diverge but the sum of them converges?

    Thank you
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  2. #2
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    To find \sum_{n=1}^{\infty}{\ln{\frac{n}{n+1}}}, we may begin with the fact that

    \sum_{n=1}^{a}{\ln\frac{n}{n+1}}=\ln{\frac{1}{2}}+  \ln{\frac{2}{3}}+\ln{\frac{3}{4}}+\cdots+\ln{\frac  {a}{a+1}}.

    How can this sum be simplified?
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  3. #3
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    Quote Originally Posted by joepinsley View Post
    Infinite
    Σ ln(n/(n+1))
    n=1

    I know that you have to test and make sure that "a sub n" goes to 0 as n goes to infinite. You could split the summation into Σln(n) - Σln(n+1) and then take the limits as n approaches infinite of each "a sub n" separately, right? You get infinite minus infinite, though. But isn't it possible that these two limits diverge but the sum of them converges?

    Thank you
    Why not try the integral test for convergence?

     \displaystyle \sum_{n = k}^{\infty} f(n) is convergent if  \int_k^{\infty} f(x) dx is finite.

    So  \int_1^{\infty} \ln(\frac{x}{x+1})dx

    Just remember that  \ln(a/b) = \ln(a) - \ln(b) and that integral is relatively easy to compute.
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  4. #4
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    Quote Originally Posted by joepinsley View Post
    Infinite
    Σ ln(n/(n+1))
    n=1

    I know that you have to test and make sure that "a sub n" goes to 0 as n goes to infinite. You could split the summation into Σln(n) - Σln(n+1) and then take the limits as n approaches infinite of each "a sub n" separately, right? You get infinite minus infinite, though. But isn't it possible that these two limits diverge but the sum of them converges?

    Thank you
    The partial sum up to n = k is given by (ln(1) - ln(2)) + (ln(2) - ln(3)) + ... + (ln(k) - ln(k+1)) = ln(1) - ln(k+1). Since ln(x) is increasing and unbounded, ln(1) - ln(k+1) approaches negative infinity as k gets bigger and bigger, so the sum diverges.
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  5. #5
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    Quote Originally Posted by rn443 View Post
    The partial sum up to n = k is given by (ln(1) - ln(2)) + (ln(2) - ln(3)) + ... + (ln(k) - ln(k+1)) = ln(1) - ln(k+1). Since ln(x) is increasing and unbounded, ln(1) - ln(k+1) approaches negative infinity as k gets bigger and bigger, so the sum diverges.
    So is it wrong to think of it as: ln(1) - ln(2) + ln(2) -ln(3) + ln(3)....
    and consequently cancel -ln(2) and ln(2), -ln(3) and ln(3), etc., leaving only ln(1)??
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  6. #6
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    Quote Originally Posted by joepinsley View Post
    So is it wrong to think of it as: ln(1) - ln(2) + ln(2) -ln(3) + ln(3)....
    and consequently cancel -ln(2) and ln(2), -ln(3) and ln(3), etc., leaving only ln(1)??
    That's wrong. It will not only leave  \ln(1) , it will also leave  -\ln(k+1)

    For example, if the series was:

     \sum_{n = 1}^2 \ln{\frac{n}{n+1}}

    You'd have:

    ln(1) - \ln(2) + \ln(2) - \ln(3)  = \ln(1) - \ln(3)

    And so for the series we have:

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

    You'd have:

     \ln(1) - \ln(\infty + 1) (if you'll excuse the unforgivable notation!)

    Which quite clearly gives the value  - \infty
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  7. #7
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    I will excuse the notation, certainly! Thank you for spelling it out so clearly. I see now. Thank you for helping me understand.
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