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Math Help - Calc 2: Converge/Diverge

  1. #1
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    Calc 2: Converge/Diverge

    How do i find out if these converge/diverge:

    inifinity
    SIGMA sqrt(2n+1)/(n^2)
    n=1

    infinity
    SIGMA (3^K+K)/(K!)
    k=1

    I dont know what K! is and i dont know what to do with the square root. It stumps me and I dont know what theorems to use. The integral test? Bounded sum test? p-series test?
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    Quote Originally Posted by MathNeedy18 View Post
    How do i find out if these converge/diverge:

    inifinity
    SIGMA sqrt(2n+1)/(n^2)
    n=1

    infinity
    SIGMA (3^K+K)/(K!)
    k=1

    I dont know what K! is and i dont know what to do with the square root. It stumps me and I dont know what theorems to use. The integral test? Bounded sum test? p-series test?
    0 < \frac{\sqrt{2n+1}}{n^2} < \frac{\sqrt{2n+2n}}{n^2} = \frac{2\sqrt{n}}{n^2} = \frac{2}{n^{3/2}} .......

    For the second one, have you tried the ratio test?
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    Thanks Mr. Fantastic. My math teacher has been zooming through chapters cuz we are weeks behind and therefore just writes down definitions and what not and doesnt do examples. I learn from examples so I'm not even quite sure how to do the ratio test
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    Quote Originally Posted by MathNeedy18 View Post
    Thanks Mr. Fantastic. My math teacher has been zooming through chapters cuz we are weeks behind and therefore just writes down definitions and what not and doesnt do examples. I learn from examples so I'm not even quite sure how to do the ratio test
    Does your book have a chapter or section that explains the ratio test and gives examples .....?
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    Hi,

    Often remember to thank Riemann ^^

    We call the ratio test "d'Alembert test" in France, if it can help...

    If the general term is An, if \frac{A_{n+1}}{A_n} < 1, the serie converges. If > 1, it diverges. If =1, you have to try
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    Quote Originally Posted by Moo View Post
    Hi,

    Often remember to thank Riemann ^^

    We call the ratio test "d'Alembert test" in France, if it can help...

    If the general term is An, if \frac{A_{n+1}}{A_n} < 1, the serie converges. If > 1, it diverges. If =1, you have to try
    The ratio test will be one of your best friends in proving the convergence or divergence of a series. Other tests that I found most useful were the comparison test, root (although it fails when ratio test fails) and limit comparison test.
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    Quote Originally Posted by colby2152 View Post
    root (although it fails when ratio test fails) and
    This is WRONG. The root test is SUPERIOR to the ratio test.

    The actual full statement is based on limsup and liminf. The test says given \sum_{n=0}^{\infty}a_n x^n. Consider L = \limsup |a_n|^{1/n}, the beauty here is that this limit exists (or L=\infty) not matter what, the reciprocal will be the radius of convergence. The ratio test is not superior, for one thing we require that the terms a_n be non-zero to use this test! While in root test this is not a necessity! Furthermore, the ratio test is based on root test, because of the fact that: \liminf |a_{n+1}/a_n| \leq L \leq \limsup |a_{n+1}/a_n|. Thus, if \limsup |a_{n+1}/a_n| < 1 then L<1 and we have convergence. If \liminf |a_{n+1}/a_n| > 1 then L>1 and we have divergence. All because of the root test. And finally we can give examples where there ratio test fails and root test succedes.
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    Quote Originally Posted by ThePerfectHacker View Post
    This is WRONG. The root test is SUPERIOR to the ratio test.

    The actual full statement is based on limsup and liminf. The test says given \sum_{n=0}^{\infty}a_n x^n. Consider L = \limsup |a_n|^{1/n}, the beauty here is that this limit exists (or L=\infty) not matter what, the reciprocal will be the radius of convergence. The ratio test is not superior, for one thing we require that the terms a_n be non-zero to use this test! While in root test this is not a necessity! Furthermore, the ratio test is based on root test, because of the fact that: \liminf |a_{n+1}/a_n| \leq L \leq \limsup |a_{n+1}/a_n|. Thus, if \limsup |a_{n+1}/a_n| < 1 then L<1 and we have convergence. If \liminf |a_{n+1}/a_n| > 1 then L>1 and we have divergence. All because of the root test. And finally we can give examples where there ratio test fails and root test succedes.
    Whoops, I got it backwards... the ratio test fails when the root test fails...
    Last edited by colby2152; August 7th 2008 at 06:49 AM.
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  9. #9
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    Never heard 'bout the superiority of one test upon another... The two have always been equivalent to us
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