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Math Help - Discuss the convergence of the series

  1. #1
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    Discuss the convergence of the series

    I have tried the ratio test and the comparison test but still i am unable to get the result

    Question : Discuss the convergence of the infinite series \sum_{n=1}^{\infty} \frac{3n}{4(n+1)}

    Solution :

    Ratio test :

     \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \lim_{n \to \infty} \frac{3(n+1)}{4(n+2)}. \frac{4(n+1)}{3n} = \lim_{n \to \infty}  \frac{(n+1)^2}{(n+2)n} now dividing nu and de by n^2 we get

    \lim_{n \to \infty} \frac{1+\frac{2}{n}+\frac{1}{n^2}}{1+\frac{2}{n}} = 1 (Ratio test failed )


    Comparison test :

    \lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \frac{3n}{4(n+1)}

    Here u_n = \frac{3n}{4(n+1)} ; v_n = \frac{1}{n}

    \lim_{n \to \infty } \frac{u_n}{v_n} = \lim_{n \to \infty } \frac{3}{4(n+1)} = \frac{3}{4} \lim_{n \to \infty }  \frac{3}{4(n+1)} dividing the nu and de by n weget

    \frac{3}{4} \lim_{n \to \infty }  \frac{\frac{3}{n}}{4+\frac{1}{n}} = \frac{3}{4} . 0 = 0

    What should i do !!!!!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flintstone View Post
    I have tried the ratio test and the comparison test but still i am unable to get the result

    Question : Discuss the convergence of the infinite series \sum_{n=1}^{\infty} \frac{3n}{4(n+1)}

    [/COLOR][/I][/B]!!!!!
    You have got to be kidding me? Try comparing the series to 1
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    Quote Originally Posted by Drexel28 View Post
    You have got to be kidding me? Try comparing the series to 1
    Sorry i didnt get u ......which test should i use !!!!!!!
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flintstone View Post
    Sorry i didnt get u ......which test should i use !!!!!!!
    Try the limit comparison test.
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    Quote Originally Posted by Drexel28 View Post
    Try the limit comparison test.
    sir i have tried the limit comparison test .........please see my first post for the steps .....
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flintstone View Post
    sir i have tried the limit comparison test .........please see my first post for the steps .....
    \lim_{n\to\infty}\frac{\frac{3n}{4(n+1)}}{1}=\frac  {3}{4}

    Look up limit test for divergence.
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    Quote Originally Posted by Drexel28 View Post
    \lim_{n\to\infty}\frac{\frac{3n}{4(n+1)}}{1}=\frac  {3}{4}

    Look up limit test for divergence.
    u have taken u_n = \frac{3n}{4(n+1)} and v_n = 1 is that correct ??........are we allowed to take v_n =1 (just 1)
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flintstone View Post
    u have taken u_n = \frac{3n}{4(n+1)} and v_n = 1 is that correct ??........are we allowed to take v_n =1 (just 1)
    Surely.
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  9. #9
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    What Drexel is saying is that in any series,

    If \lim_{n \to \infty}{a_n} \neq 0,

    then the series diverges.


    Look at the n^{th} term in your series.

    a_n = \frac{3n}{4(n + 1)}

    \lim_{n \to \infty}a_n = \lim_{n \to \infty}\frac{3n}{4(n + 1)}

     = \frac{3}{4}\lim_{n \to \infty}\frac{n}{n + 1}

     = \frac{3}{4}\lim_{n \to \infty}\left(1 - \frac{1}{n + 1}\right)

     = \frac{3}{4}\cdot 1

     = \frac{3}{4}.


    Since this is not 0, the series diverges.
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