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Math Help - Determining if a series is convergent or divergent

  1. #1
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    Determining if a series is convergent or divergent

    Hi I was wondering if anyone could explain this to me,

    I have

    Series (-1)^n n/ln(n)
    sum : infinite
    n:2

    I took the derivative and got : 1/ln(x) - 1/(ln(x)^2)

    How do I determine if this is a decreasing or increasing function.

    From what I could tell 2 is the only value to make it go <0 , so is this neither increasing or decreasing?
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  2. #2
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    Quote Originally Posted by illidari View Post
    Hi I was wondering if anyone could explain this to me,

    I have

    Series (-1)^n n/ln(n)
    sum : infinite
    n:2

    I took the derivative and got : 1/ln(x) - 1/(ln(x)^2)

    How do I determine if this is a decreasing or increasing function.

    From what I could tell 2 is the only value to make it go <0 , so is this neither increasing or decreasing?

    You wrote \sum\limits^\infty_{n=2}(-1)^n\frac{n}{\ln n} . This series diverges since the general term' sequence doesn't converge to zero.
    I can't understand what you're trying to derivate and wrt what...

    Tonio
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  3. #3
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    Yeah that is the series I wrote, I was trying to apply the alternating series test. The 2nd part of the equation, excluding (-1)^n, needs to be decreasing

    I took the derivative of it and need to determine if that derivative is <0 .

    That is the part I got stuck :/
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  4. #4
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    Quote Originally Posted by illidari View Post
    Yeah that is the series I wrote, I was trying to apply the alternating series test. The 2nd part of the equation, excluding (-1)^n, needs to be decreasing

    I took the derivative of it and need to determine if that derivative is <0 .

    That is the part I got stuck :/

    Are you trying to derivate wrt n, which is a discrete variable?! It can't be done, of course. Besides this the series does
    not converge at all, Leibnitz or not.

    Tonio
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