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Math Help - integral test

  1. #1
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    Wink integral test

    I just want to quickly check....

    I used the integral test to check if \sum_{n=2}^{\infty}\frac{1}{nln(n)} converges.... and i found that it does. Can someone verify this? Thanks
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  2. #2
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    Hi,
    Quote Originally Posted by tsal15 View Post
    I used the integral test to check if \sum_{n=2}^{\infty}\frac{1}{nln(n)} converges.... and i found that it does.
    No it does not, this series is divergent. (it can be shown using the integral test )
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  3. #3
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    Quote Originally Posted by tsal15 View Post
    I just want to quickly check....

    I used the integral test to check if \sum_{n=2}^{\infty}\frac{1}{nln(n)} converges.... and i found that it does. Can someone verify this? Thanks
    \sum_{n=2}^{\infty}\frac{1}{n \ln(n)} > \int_2^{+\infty} \frac{1}{x \ln x} \, dx and I'm pretty confident that the integral does not have a finite value .....
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    Quote Originally Posted by mr fantastic View Post
    \sum_{n=2}^{\infty}\frac{1}{n \ln(n)} > \int_2^{+\infty} \frac{1}{x \ln x} \, dx and I'm pretty confident that the integral does not have a finite value .....
    Are you sure that sign is around the right way?
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Are you sure that sign is around the right way?
    All the left hand rectangles of width 1 (starting from x = 2) have areas lying above the corresponding area under the curve so I'm pretty sure I'm sure.

    But since I'm about to go to bed I will leave it to others to point out my mistakes if there are any.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    All the left hand rectangles of width 1 (starting from x = 2) have areas lying above the corresponding area under the curve so I'm pretty sure I'm sure.

    But since I'm about to go to bed I will leave it to others to point out my mistakes if there are any.
    Yes that would make sense.

    Plus, to show that it's divergent it WOULD have to be greater than something tending to infinity wouldn't it...
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  7. #7
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    Always verify that f must be positive, continuous and decreasing on [1,\infty[.
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    Quote Originally Posted by Krizalid View Post
    Always verify that f must be positive, continuous and decreasing on [1,\infty[.
    Yes when using the integral test

    Thanks for the reminder
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    \sum_{n=2}^{\infty}\frac{1}{n \ln(n)} > \int_2^{+\infty} \frac{1}{x \ln x} \, dx and I'm pretty confident that the integral does not have a finite value .....
    Ok now, I've got a problem...

    the integral equals => ln(ln(x) with infinity and 2 being the bounds

    therefore, ln(ln(\infty)) - ln(ln(2))

    ln(ln(2)) is a value. Thus even if ln(ln(\infty)) does diverge, ln(ln(2)) will converge..... so does this change things?
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  10. #10
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    Quote Originally Posted by tsal15 View Post
    Ok now, I've got a problem...

    the integral equals => ln(ln(x) with infinity and 2 being the bounds

    therefore, ln(ln(\infty)) - ln(ln(2))

    ln(ln(2)) is a value. Thus even if ln(ln(\infty)) does diverge, ln(ln(2)) will converge..... so does this change things?
    Nope, the integral diverges.
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  11. #11
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    Quote Originally Posted by tsal15 View Post
    Ok now, I've got a problem...

    the integral equals => ln(ln(x) with infinity and 2 being the bounds

    therefore, ln(ln(\infty)) - ln(ln(2))

    ln(ln(2)) is a value. Thus even if ln(ln(\infty)) does diverge, ln(ln(2)) will converge..... so does this change things?
    You're interested in the final value of the integral, not the individual bits that contribute to calculating this value.
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  12. #12
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    Quote Originally Posted by tsal15 View Post
    Ok now, I've got a problem...

    the integral equals => ln(ln(x) with infinity and 2 being the bounds

    therefore, ln(ln(\infty)) - ln(ln(2))

    ln(ln(2)) is a value. Thus even if ln(ln(\infty)) does diverge, ln(ln(2)) will converge..... so does this change things?
    In the grand scheme of things, \ln{\ln{2}} is very small when compared to infinity.

    So even though this value is being subtracted, it's still close enough to infinity to say that the integral tends to infinity.

    Therefore it diverges.

    Make sense?
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