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Math Help - Test the convergence of the series

  1. #1
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    Test the convergence of the series

    Question : Test the convergence of the series \sum_{n = 1}^\infty  \frac{3n}{2(n + 1)}
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  2. #2
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    use the divergence test to show that the limit as n->infinity =3/2 so it diverges
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  3. #3
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    Can u please provide me with the starting few steps

    Quote Originally Posted by Sam1111 View Post
    use the divergence test to show that the limit as n->infinity =3/2 so it diverges

    Can u please provide me with the starting few steps
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  4. #4
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    The divergence test states that if the limit of 3n/(2(n+1)) doesnt = 0 then the series diverges.

    So take the limit of 3n/(2(n+1)) by dividing everything by the highest power in the denominator. Then evaluate this letting n->infinity.
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  5. #5
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    Could u please tell me if its correct or no

    Quote Originally Posted by Sam1111 View Post
    The divergence test states that if the limit of 3n/(2(n+1)) doesnt = 0 then the series diverges.

    So take the limit of 3n/(2(n+1)) by dividing everything by the highest power in the denominator. Then evaluate this letting n->infinity.

    I am not so good with the convergent / divergent series so i dont know whether i have done correctly or no please check and let me know

    \sum_{n=1}^{ \infty} \frac{3n}{2(n+1)}

    \lim_{n \to infty} \frac{3n}{2(n+1)} = \lim_{n \to infty} \frac{3n}{n (2+ \frac{1}{n})} = \lim_{n \to \infty} \frac{3}{2 + \frac{1}{n}} = \frac{3}{2}

    since \sum_{n = 1}^{ \infty} u_n \ne 0 then series is divergent

    \therefore \lim_{n \to \infty} \frac{3n}{2(n+1)} is divergent
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  6. #6
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    Quote Originally Posted by zorro View Post
    I am not so good with the convergent / divergent series so i dont know whether i have done correctly or no please check and let me know

    \sum_{n=1}^{ \infty} \frac{3n}{2(n+1)}

    \lim_{n \to infty} \frac{3n}{2(n+1)} = \lim_{n \to infty} \frac{3n}{n (2+ \frac{1}{n})} = \lim_{n \to \infty} \frac{3}{2 + \frac{1}{n}} = \frac{3}{2}

    since \sum_{n = 1}^{ \infty} u_n \ne 0 then series is divergent

    \therefore \lim_{n \to \infty} \frac{3n}{2(n+1)} is divergent
    No no no ...

    A series \sum_{i=1}^{\infty} a_i converges only if \lim_{i \to \infty}a_i=0. In your case a_i \to 3/2 \ne 0.

    CB
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  7. #7
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    I didnt get u

    Quote Originally Posted by CaptainBlack View Post
    No no no ...

    A series \sum_{i=1}^{\infty} a_i converges only if \lim_{i \to \infty}a_i=0. In your case a_i \to 3/2 \ne 0.

    CB

    Thats what i am saying the series is divergent ...........since \sum_{n = 1}^{ \infty} u_n \ne 0

    U got me there mite i am confused now
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  8. #8
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    Quote Originally Posted by zorro View Post
    Thats what i am saying the series is divergent ...........since \sum_{n = 1}^{ \infty} u_n \ne 0

    U got me there mite i am confused now
    No since it's divergent because \lim_{n \to \infty}u_n \ne 0 which is not the same thing as you posted.

    To see this consider:

    \sum_{n = 1}^{ \infty} 0.1^n=\frac{1}{9}

    So if u_n=0.1^n we have \sum_{n = 1}^{ \infty} u_n \ne 0 and yet the series converges, but we do have \lim_{n \to \infty}u_n = 0 since this is a necessary condition for the series to converge.

    CB
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  9. #9
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    Quote Originally Posted by CaptainBlack View Post
    No since it's divergent because \lim_{n \to \infty}u_n \ne 0 which is not the same thing as you posted.

    To see this consider:

    \sum_{n = 1}^{ \infty} 0.1^n=\frac{1}{9}

    So if u_n=0.1^n we have \sum_{n = 1}^{ \infty} u_n \ne 0 and yet the series converges, but we do have \lim_{n \to \infty}u_n = 0 since this is a necessary condition for the series to converge.

    CB

    So u mean to say i should have posted as \lim_{n \to \infty}u_n \ne 0 that is why the series is divergent is tht correct
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by zorro View Post
    So u mean to say i should have posted as \lim_{n \to \infty}u_n \ne 0 that is why the series is divergent is tht correct
    Yes.

    CB
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