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Thread: Using integrals :D

  1. #1
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    Using integrals :D

    $\displaystyle {{\lim_{\substack{n\rightarrow\infty}} {\left( \frac {1}{\sqrt{n}\sqrt{n+1}}+\frac {1}{\sqrt{n}\sqrt{n+2}}+...+\frac {1}{\sqrt{n}\sqrt{n+n}}\right)}$
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by rondo09 View Post
    $\displaystyle {{\lim_{\substack{n\rightarrow\infty}} {\left( \frac {1}{\sqrt{n}\sqrt{n+1}}+\frac {1}{\sqrt{n}\sqrt{n+2}}+...+\frac {1}{\sqrt{n}\sqrt{n+n}}\right)}$
    $\displaystyle {\lim_{n\rightarrow\infty}( \frac {1}{\sqrt{n}\sqrt{n+1}}+\frac {1}{\sqrt{n}\sqrt{n+2}}+...+\frac {1}{\sqrt{n}\sqrt{n+n}})$

    = $\displaystyle {\lim_{n\rightarrow\infty}\sum_{i=1}^{n}( \frac {1}{\sqrt{n}\sqrt{n+i}})$

    = $\displaystyle {\lim_{n\rightarrow\infty}\sum_{i=1}^{n}(\frac{1}{ \sqrt{1+\frac{i}{n}}})(\frac{1}{n})$

    = $\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1+x}}$

    = $\displaystyle 2\sqrt{1+x}\big|_0^{1}$

    = $\displaystyle 2(\sqrt{2}-1)$
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