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Math Help - Root and Ratio Tests - Testing for divergence or convergence

  1. #1
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    Root and Ratio Tests - Testing for divergence or convergence

    \sum_{n=1}^{\infty}{\frac{(n+1)^{2n}}{2^nn^{2n}}}

    \sum_{n=2}^{\infty}{({{{\frac{n}{n+1}}})^{n^{2}}}}

    I have tried both tests on the series, and both yield the limit going to 1 (test fails, may diverge or converge).

    Is there a way to show either diverge or converge by another test, or comparison theorem? Can't seem to find anything! Will keep trying though!

    Thank you!!
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  2. #2
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    There is the integral test.

    That could prove messy, though.

    Integral test for convergence - Wikipedia, the free encyclopedia
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  3. #3
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    Quote Originally Posted by matt.qmar View Post
    \sum_{n=1}^{\infty}{\frac{(n+1)^{2n}}{2^nn^{2n}}}

    \sum_{n=2}^{\infty}{({{{\frac{n}{n+1}}})^{n^{2}}}}

    I have tried both tests on the series, and both yield the limit going to 1 (test fails, may diverge or converge).

    Is there a way to show either diverge or converge by another test, or comparison theorem? Can't seem to find anything! Will keep trying though!

    Thank you!!

    I don't get it: in both series I get something very nice and less than one after applying the root test:

    \sqrt[n]{\frac{\left(n+1\right)^{2n}}{2^nn^{2n}}}=\frac{\l  eft(n+1\right)^2}{2n^2} \xrightarrow [n \to \infty]{}\frac{1}{2}<1

    \sqrt[n]{\left(\frac{n}{n+1}\right)^{n^2}}=\left(\frac{n}{  n+1}\right)^n=\frac{1}{\left(1+\frac{1}{n}\right)^  n}\xrightarrow [n \to \infty]{}\frac{1}{e}

    Tonio
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  4. #4
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    Quote Originally Posted by tonio View Post

    \sqrt[n]{\left(\frac{n}{n+1}\right)^{n^2}}=\left(\frac{n}{  n+1}\right)^n=\frac{1}{\left(1+\frac{1}{n}\right)^  n}\xrightarrow [n \to \infty]{}\frac{1}{e}
    Why soes this tend to {1 \over e} ?
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  5. #5
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    Quote Originally Posted by bigdoggy View Post
    Why soes this tend to {1 \over e} ?

    A very well know limit: \left(1+\frac{1}{n}\right)^n \xrightarrow [n\to \infty] {}e

    Tonio
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