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Math Help - converging or diverging?

  1. #1
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    Wink converging or diverging?

    my friend and i are having differences on this problem .... please help

    sumation of n=1 to inifinity ((n-1)/(n+1))^n(n+1)

    he keeps getting that series diverges due to the divergence test
    I keep getting that the series absolutely converges due to the root test

    please help and show us who is right
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  2. #2
    Math Engineering Student
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    well, applying the root test we have to compute \underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{n-1}{n+1} \right)^{n+1}=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{1-\frac{1}{n}}{1+\frac{1}{n}} \right)^{n}\left( \frac{n-1}{n+1} \right)=\frac{e^{-1}}{e}=\frac{1}{e^{2}}<1.

    so the series converges, and absolutely, since it's a series with positive terms.
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  3. #3
    Eater of Worlds
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    \sum_{n=1}^{\infty}\left(\frac{n-1}{n+1}\right)^{n(n+1)}

    Did you try the ratio test?. By the ratio test, we get {\rho}=e^{-2}

    Which means it converges, since {\rho}<1

    \sum_{n=1}^{\infty}\left(\frac{n-1}{n+1}\right)^{n(n+1)}=\frac{1351}{814589}
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  4. #4
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    thank you! that helped in solving a debate we were having.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    A more highfalutan to do it is to notice that 0<\left(\frac{n-1}{n+1}\right)^{n(n+1)}<\left(\frac{n}{n+1}\right)  ^{n^2}, and with a little algebraic manipulation, the RHS is equal to e^{n^2\ln\left(1-\frac{1}{n+1}\right)}. Notice though that \ln\left(1-\frac{1}{n+1}\right)\stackrel{n\to\infty}\sim \frac{-1}{n+1}\stackrel{n\to\infty}{\sim}\frac{-1}{n} which of course implies that e^{n^2\ln\left(1-\frac{1}{n+1}\right)}\stackrel{n\to\infty}{\sim}e^  {-n}. So then we can see that \sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^2  } converges by the limit comparison test with \sum_{n=1}^{\infty}e^{-n}, which then means that the actual series converes by the comparison test.
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