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Math Help - Test the Convergence of the series

  1. #1
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    Test the Convergence of the series

    This question is from my exam paper which i had appeared last semister i am trying to solve this but dont know the right answer.

    Q: Test the convergence of the series


    ∑ 3n/2(n+1)
    n=1


    Please replies with explanation if possible .......
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  2. #2
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    Quote Originally Posted by varunnayudu View Post
    This question is from my exam paper which i had appeared last semister i am trying to solve this but dont know the right answer.

    Q: Test the convergence of the series


    ∑ 3n/2(n+1)
    n=1


    Please replies with explanation if possible .......
    Hint: A necessary (but not sufficient) condition for the convergence of \sum_{n = 1}^{\infty} a_n is that \lim_{n \rightarrow \infty} a_n = 0.
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  3. #3
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    Please can u evaluate this limit

    Evaluate
    lim 3n/2(n+1)
    n→∞

    I have evaluated it but dont know if its right or wrong...
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  4. #4
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    Quote Originally Posted by varunnayudu View Post
    Evaluate
    lim 3n/2(n+1)
    n→∞

    I have evaluated it but dont know if its right or wrong...
    Take out a factor of n on both sides:

     \frac{n}{n} . \frac{3}{2+\frac{1}{n}} = \frac{3}{2+\frac{1}{n}}

    Is that easier?
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  5. #5
    Member Greengoblin's Avatar
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    Or you can use l'Hopital's rule.
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  6. #6
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    Direct comparison test also works fine:

    \frac{3n}{2n+2}\ge\frac{3n}{2n^{2}+2n^{2}}=\frac{3  n}{4n^{2}}=\frac34\cdot\frac{1}{n}.

    Hence, your series diverges.
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by varunnayudu View Post
    Test the convergence of the series \sum_{n=1}^\infty \frac{3n}{2(n+1)}

    Note: Please specify which test to use ,Ratio Test,Integral Test,Raabe Test etc.....
    There is probably one that is a little easier to use, but you can use the integral test here:

    \sum_{n=1}^{\infty}\frac{3n}{2\left(n+1\right)}\im  plies\tfrac{3}{2}\int_1^{\infty}\frac{x}{x+1}\,dx

    Let z=x+1\implies x=z-1. Thus, \,dz=\,dx

    Thus, the integral becomes \tfrac{3}{2}\int_2^{\infty}\frac{z-1}{z}\,dz=\tfrac{3}{2}\int_2^{\infty}\left(z-\frac{1}{z}\right)\,dz

    This gives us \tfrac{3}{2}\left.\left[\tfrac{1}{2}z^2-\ln\left|z\right|\right]\right|_2^\infty=\infty

    Thus, this diverges by the integral test.

    -------------------------------------------------------------------------

    I just thought of an easier way to do this...

    You can show that the series is divergent by the nth term test (also called Divergence Test):

    \lim_{n\to\infty}\frac{3n}{2\left(n+1\right)}=\fra  c{3}{2}\neq0. Thus, the series diverges.
    Last edited by mr fantastic; December 23rd 2008 at 07:25 PM. Reason: No edit - just flagging this reply as having been moved from a double post
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  8. #8
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    Quote Originally Posted by Chris L T521 View Post
    There is probably one that is a little easier to use, but you can use the integral test here:



    \sum_{n=1}^{\infty}\frac{3n}{2\left(n+1\right)}\im  plies\tfrac{3}{2}\int_1^{\infty}\frac{x}{x+1}\,dx



    Let z=x+1\implies x=z-1. Thus, \,dz=\,dx



    Thus, the integral becomes \tfrac{3}{2}\int_2^{\infty}\frac{z-1}{z}\,dz=\tfrac{3}{2}\int_2^{\infty}\left(z-\frac{1}{z}\right)\,dz



    This gives us \tfrac{3}{2}\left.\left[\tfrac{1}{2}z^2-\ln\left|z\right|\right]\right|_2^\infty=\infty



    Thus, this diverges by the integral test.

    [snip]
    Integral test requires that f must be positive, continuous and decreasing on [1,\infty[ then it applies.

    Be sure to verify these conditions before you may apply it.
    Last edited by mr fantastic; December 23rd 2008 at 07:24 PM. Reason: Added the quote of Chris's post. Note also that this reply has been moved from a double post
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