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Math Help - nth root of (n + sqrt(n))

  1. #1
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    nth root of (n + sqrt(n))

    I have to show that \lim\sqrt[n]{n+\sqrt{n}} goes to 1 as n\rightarrow\infty.
    I have that \sqrt[n]{n}\leq \sqrt[n]{n+\sqrt{n}}\leq\sqrt[n]{2n}
    I know that \sqrt[n]{n}\rightarrow 1 via the binomial theorem.
    If I can show that \sqrt[n]{2n}\rightarrow 1, I could use the Squeeze theorem.
    Any ideas?
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  2. #2
    Junior Member qspeechc's Avatar
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    Quote Originally Posted by dannyboycurtis View Post
    I have to show that \lim\sqrt[n]{n+\sqrt{n}} goes to 1 as n\rightarrow\infty.
    I have that \sqrt[n]{n}\leq \sqrt[n]{n+\sqrt{n}}\leq\sqrt[n]{2n}
    I know that \sqrt[n]{n}\rightarrow 1 via the binomial theorem.
    If I can show that \sqrt[n]{2n}\rightarrow 1, I could use the Squeeze theorem.
    Any ideas?
    If you know \sqrt[n]{n}\rightarrow 1 doesn't that help you with \sqrt[n]{2n}\rightarrow 1? If you can show \sqrt[n]{2}\rightarrow 1 you'll be done right? You will use the product of convergent sequences theorem.
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  3. #3
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    Quote Originally Posted by dannyboycurtis View Post
    I have to show that \lim\sqrt[n]{n+\sqrt{n}} goes to 1 as n\rightarrow\infty.
    I have that \sqrt[n]{n}\leq \sqrt[n]{n+\sqrt{n}}\leq\sqrt[n]{2n}
    I know that \sqrt[n]{n}\rightarrow 1 via the binomial theorem.
    If I can show that \sqrt[n]{2n}\rightarrow 1, I could use the Squeeze theorem.
    Any ideas? Yes: \sqrt[n]{{2n}} = \sqrt[n]{2}\sqrt[n]{n}.
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