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Math Help - testing series for conv/div.

  1. #1
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    testing series for conv/div.

    how do i find if a series converges/diverges for this type of series:

    inf
    Σ [(-1)^k]*[(2^k)/ln (k)]
    k=2
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  2. #2
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    Solution

    I'm wondering if you could use the ratio test for this:
    |an+1/an|, So you would find the an+1 term:

    [(-1)^(k+1)]*(2^(k+1))/ln(k+1)

    Now, do the ratio:
    [[(-1)^(k+1)]*(2^(k+1))/ln(k+1)] / [((-1)^k)*(2^k)/ln(k)]

    Can be rewritten as:
    [[(-1)^(k)*(-1)]*(2^(k)*2))*ln(k)] / [ln(k+1)((-1)^k)*(2^k)]

    Now, you can cancel the (-1)^k term and the 2^k term from the numerator and the denominator to get:
    (-1)*2*ln(k) / (ln(k+1))

    Remember, you take the absolute value of the ratio so it reduces to:

    2lnk/ln(k+1)

    Now I don't know how to show that as k --> infinity, ln(k) and ln(k+1) are essentially equal. Then, you would be left with a ratio of 2. Since 2>1, the series would be said to diverge. I don't really know if this helps or not.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by VorpalX7 View Post
    how do i find if a series converges/diverges for this type of series:

    inf
    Σ [(-1)^k]*[(2^k)/ln (k)]
    k=2
    If this is \sum_{n=2}^{\infty}\frac{(-1)^n2^n}{\ln(n)} consider that \lim_{n\to\infty}\frac{(-1)^n2^n}{\ln(n)}\ne{0} therefore the series diverges by the n-th term test
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