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Math Help - test for convergence

  1. #1
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    test for convergence

    test for convergence: sigma from n=1 to infinitie of [1-(1/squarerootn)]^n
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  2. #2
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    screams comparison test to me. think famous limits. however im pretty rubbish at analysis.
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  3. #3
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    Quote Originally Posted by twilightstr View Post

    test for convergence: \sum_{n=1}^{\infty} \left(1- \frac{1}{\sqrt{n}} \right)^n
    applying the mean value theorem to the function f(x)=\ln x, \ 0 < a < b, we'll get: \ln b - \ln a > \frac{b-a}{b}. \ \ \ \ \ (1)

    now suppose n \geq 2 and in (1) put a=\sqrt{n} - 1 and b=\sqrt{n} to get \ln \sqrt{n} - \ln(\sqrt{n} - 1) > \frac{1}{\sqrt{n}}. therefore:

    \left(1 - \frac{1}{\sqrt{n}} \right)^n < e^{-\sqrt{n}} and hence your series is convergent by the comparison test.

    (note that \lim_{n\to\infty} \frac{e^{-\sqrt{n}}}{\frac{1}{n^2}}=\lim_{x\to\infty} \frac{x^4}{e^x}=0 and thus the series \sum e^{-\sqrt{n}} is convergent by the limit comparison test.)
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    applying the mean value theorem to the function f(x)=\ln x, \ 0 < a < b, we'll get: \ln b - \ln a > \frac{b-a}{b}. \ \ \ \ \ (1)

    now suppose n \geq 2 and in (1) put a=\sqrt{n} - 1 and b=\sqrt{n} to get \ln \sqrt{n} - \ln(\sqrt{n} - 1) > \frac{1}{\sqrt{n}}. therefore:

    \left(1 - \frac{1}{\sqrt{n}} \right)^n < e^{-\sqrt{n}} and hence your series is convergent by the comparison test.

    (note that \lim_{n\to\infty} \frac{e^{-\sqrt{n}}}{\frac{1}{n^2}}=\lim_{x\to\infty} \frac{x^4}{e^x}=0 and thus the series \sum e^{-\sqrt{n}} is convergent by the limit comparison test.)
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    applying the mean value theorem to the function f(x)=\ln x, \ 0 < a < b, we'll get: \ln b - \ln a > \frac{b-a}{b}. \ \ \ \ \ (1)

    now suppose n \geq 2 and in (1) put a=\sqrt{n} - 1 and b=\sqrt{n} to get \ln \sqrt{n} - \ln(\sqrt{n} - 1) > \frac{1}{\sqrt{n}}. therefore:

    \left(1 - \frac{1}{\sqrt{n}} \right)^n < e^{-\sqrt{n}} and hence your series is convergent by the comparison test.
    Just a quick remark: we can also see this as the usual convexity inequality 1-x\leq e^{-x} for x=\frac{1}{\sqrt{n}}, then raised to the power n.
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