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Math Help - Simple Question

  1. #1
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    Simple Question

    sn = (((-1)^n)*n)/(2n - 1)

    How would I determine the convergence or divergence of the sequence (sn) and find the limit if it exists?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bearej50 View Post
    sn = (((-1)^n)*n)/(2n - 1)

    How would I determine the convergence or divergence of the sequence (sn) and find the limit if it exists?
    As this sequence has alternating signs it can only converge to 0 and so |s_n| must also converge to 0. For n>0:

    |s_n|=n/(2n-1)=1/(2-1/n) \to 1/2 as n \to \infty

    so the original sequence does not converge.

    CB
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  3. #3
    Super Member Showcase_22's Avatar
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    S_n=\frac{n(-1)^n}{2n-1}
    I would see if this has a limit by the ratio test:

    \frac{S_{n+1}}{S_n}=\frac{(n+1)(-1)^{n+1}}{2(n+1)-1}. \frac{2n-1}{n(-1)^n}=\frac{-(n+1)(2n-1)}{n(2n+1)}=-(1+\frac{1}{n})(\frac{2n-1}{2n+1})

    =-(1+\frac{1}{n})(\frac{2n+1-2}{2n+1})=-(1+\frac{1}{n})(1-\frac{2}{2n+1})=-(1+\frac{1}{n}-\frac{2}{2n+1}-\frac{2}{n(2n+1)}) \rightarrow -1

    but for the ratio test (I should have probably put this in before =S) we need to take the modulus.

    |\frac{S_{n+1}}{S_n}| \rightarrow 1

    Unfortunately, if the ratio test produces 1 the test doesn't hold. =S
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