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Thread: 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 $\displaystyle \sum_{n=1}^{\infty} \frac{3n}{4(n+1)}$

    Solution :

    Ratio test :

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

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


    Comparison test :

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

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

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

    $\displaystyle \frac{3}{4} \lim_{n \to \infty } \frac{\frac{3}{n}}{4+\frac{1}{n}}$ = $\displaystyle \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 $\displaystyle \sum_{n=1}^{\infty} \frac{3n}{4(n+1)}$

    [/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 .....
    $\displaystyle \lim_{n\to\infty}\frac{\frac{3n}{4(n+1)}}{1}=\frac {3}{4}$

    Look up limit test for divergence.
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  7. #7
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    Quote Originally Posted by Drexel28 View Post
    $\displaystyle \lim_{n\to\infty}\frac{\frac{3n}{4(n+1)}}{1}=\frac {3}{4}$

    Look up limit test for divergence.
    u have taken $\displaystyle u_n = \frac{3n}{4(n+1)}$ and $\displaystyle v_n = 1$ is that correct ??........are we allowed to take $\displaystyle 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 $\displaystyle u_n = \frac{3n}{4(n+1)}$ and $\displaystyle v_n = 1$ is that correct ??........are we allowed to take $\displaystyle 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 $\displaystyle \lim_{n \to \infty}{a_n} \neq 0$,

    then the series diverges.


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

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

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

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

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

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

    $\displaystyle = \frac{3}{4}$.


    Since this is not 0, the series diverges.
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