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Math Help - Assistance with convergence or divergence of Sequence

  1. #1
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    Assistance with convergence or divergence of Sequence

    Hey people, first post here?

    I am stuck on a homework problem regarding sequences. Basically: Determine if the sequence {an} converges, and if it does, find its limit when the integral is 1/6x+8 from (n to 8n).

    To start off with, I used u-substitution to get 1/6 x ln(u), but the limits of n to 8n are really throwing me off. Also our professor kind of rushed through the section on sequences without much examples. Any feed back is appreciated.

    Thanks
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    Re: Assistance with convergence or divergence of Sequence

    Do you mind posting the whole problem as stated in your homework? The way you've written it is kind of vague.

    Also, what "u" did you use and why?
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  3. #3
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    Re: Assistance with convergence or divergence of Sequence

    That is all the problem says. Basically: an= the integral (from n to 8n) of 1/6x + 8. From first glance it looks like an ln problem, so my u=6x+8, and dx=1/6 du.

    So I got 1/6 x ln(u), which is 1/6 x ln(6x+8). But the values of n to 8n are throwing me off. How would I find the limit to determine whether it is converging or diverging?

    Also the answer choices are (online hw--quest): limit = 1/6 ln8, 1/6 ln4/3, sequence diverges, 1/6 ln1/8, ln8 (If these help)
    Last edited by Beevo; August 31st 2012 at 11:06 AM.
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    Re: Assistance with convergence or divergence of Sequence

    Quote Originally Posted by Beevo View Post
    That is all the problem says. Basically: an= the integral (from n to 8n) of 1/6x + 8. From first glance it looks like an ln problem, so my u=6x+8, and dx=1/6 du.
    So I got 1/6 x ln(u), which is 1/6 x ln(6x+8). But the values of n to 8n are throwing me off. How would I find the limit to determine whether it is converging or diverging?
    Also the answer choices are (online hw--quest): limit = 1/6 ln8, 1/6 ln4/3, sequence diverges, 1/6 ln1/8, ln8 (If these help)
    So {a_n} = \frac{1}{6}\left( {\ln (48n + 8) - \ln (6n + 8)} \right) = \frac{1}{6}\ln \left( {\frac{{48n + 8}}{{6n + 8}}} \right) \to \frac{1}{6}\ln (8)
    Last edited by Plato; August 31st 2012 at 12:40 PM.
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    Re: Assistance with convergence or divergence of Sequence

    Quote Originally Posted by Plato View Post
    So {a_n} = \frac{1}{6}\left( {\ln (48n + 8) - \ln (6n + 8)} \right) - \frac{1}{6}\ln \left( {\frac{{48n + 8}}{{6n + 8}}} \right) \to \frac{1}{6}\ln (8)
    Thanks man, appreciate it. I actually got to the point of 1/6 ln(48n+8/6n+8), but did not know how to simplify it from there. I need to go back and review the rules of logarithms.
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    Re: Assistance with convergence or divergence of Sequence

    Quote Originally Posted by Beevo View Post
    Thanks man, appreciate it. I actually got to the point of 1/6 ln(48n+8/6n+8), but did not know how to simplify it from there. I need to go back and review the rules of logarithms.
    Here you are actually not using any rules specific to the logarithm. You are just taking the limit of what's inside, that is

    \lim_{n\to \infty} \frac{48n+8}{6n+8} = \lim_{n\to \infty} \frac{48+8/n}{6+8/n} = \frac{48+0}{6+0} = 8

    Since this is greater than 0 you actually have that

    \lim_{n\to\infty} \frac{1}{6}\ln\left( \frac{48n+8}{6n+8}\right) =  \frac{1}{6}\ln\left( \lim_{n\to\infty}\frac{48n+8}{6n+8}\right) = \frac{1}{6}\ln 8
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  7. #7
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    Re: Assistance with convergence or divergence of Sequence

    Quote Originally Posted by Vlasev View Post
    Here you are actually not using any rules specific to the logarithm. You are just taking the limit of what's inside, that is

    \lim_{n\to \infty} \frac{48n+8}{6n+8} = \lim_{n\to \infty} \frac{48+8/n}{6+8/n} = \frac{48+0}{6+0} = 8

    Since this is greater than 0 you actually have that

    \lim_{n\to\infty} \frac{1}{6}\ln\left( \frac{48n+8}{6n+8}\right) =  \frac{1}{6}\ln\left( \lim_{n\to\infty}\frac{48n+8}{6n+8}\right) = \frac{1}{6}\ln 8
    Yeah I eventually realized that. Guess that explains why I was initially confused. Thanks man.
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