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Math Help - Series, integral test !

  1. #1
    Junior Member xterminal01's Avatar
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    Series, integral test !

    <br />
\sum_{n=1}^\infty \frac{2}{sqrt (n^2+3)}<br />
    I tryed to use the integral test
    came up with 2/2(n^2+3)^3/2) from 1 to infinity
    0-1/8 = -1/8 ?
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by xterminal01 View Post
    <br />
\sum_{n=1}^\infty \frac{2}{sqrt (n^2+3)}<br />
    I tryed to use the integral test
    came up with 2/2(n^2+3)^3/2) from 1 to infinity
    0-1/8 = -1/8 ?
    Use the comparison test. Note that \frac{2}{\sqrt{n^2 + 3}} \geq \frac{2}{\sqrt{n^2 + 3n^2}} = \frac{1}{n} for n \geq 1.
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