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Math Help - convergency test of ((2n+1)^n)/(n^(2n))

  1. #1
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    convergency test of ((2n+1)^n)/(n^(2n))

    Hy,
    It's suppose to be simple one, but I really dont know how to start and what test should I apply.
    ((2n+1)^n)/(n^(2n))

    Should I work on ((2n+1)/(n^2))^n?
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  2. #2
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    Re: convergency test of ((2n+1)^n)/(n^(2n))

    ratio and nth root tests

    scroll down to the nth root test section + examples
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  3. #3
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    Re: convergency test of ((2n+1)^n)/(n^(2n))

    Thanx
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  4. #4
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    Re: convergency test of ((2n+1)^n)/(n^(2n))

    The ratio test also works.
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    Re: convergency test of ((2n+1)^n)/(n^(2n))

    Quote Originally Posted by Prove It View Post
    The ratio test also works.
    I have tried the ratio test, but putting n+1 instead of every n made it so complicated and stuff didn't cancel to me, I probably do it wrong.
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    Re: convergency test of ((2n+1)^n)/(n^(2n))

    Quote Originally Posted by shaharg View Post
    I have tried the ratio test, but putting n+1 instead of every n made it so complicated and stuff didn't cancel to me, I probably do it wrong.
    The root test is really simple.
    \sqrt[n]{{\frac{{{{\left( {2n + 1} \right)}^n}}}{{{n^{2n}}}}}} = \frac{{2n + 1}}{{{n^2}}} \to 0
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  7. #7
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    Re: convergency test of ((2n+1)^n)/(n^(2n))

    Quote Originally Posted by Plato View Post
    The root test is really simple.
    \sqrt[n]{{\frac{{{{\left( {2n + 1} \right)}^n}}}{{{n^{2n}}}}}} = \frac{{2n + 1}}{{{n^2}}} \to 0
    Yes it is simple.
    I appreciate your effort. Thanx a lot.
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