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Thread: infinite series

  1. December 12th 2012, 05:45 AM #1
    muddywaters
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    infinite series

    finding the general term for \frac{1}{\sqrt{2}+2}+\frac{1}{2\sqrt{3}+3\sqrt{2}} +...+\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}+...

    answer is:

    a_n=\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}=\frac{(n+1 )\sqrt{n}-n\sqrt{n+1}}{n(n+1)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}


    how to derive \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}} from \frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)}?
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  2. December 12th 2012, 05:49 AM #2
    MarkFL
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    Re: infinite series

    Use \frac{a+b}{c}=\frac{a}{c}+\frac{b}{c} then reduce.
    Thanks from muddywaters
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