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Math Help - Testing Series for Convergence/Divergence

  1. #1
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    Testing Series for Convergence/Divergence

    Hey everyone, I'm doing some practice problems to study for my calculus exam and can't seem to figure this one out. What am I missing?

    sum from n=1 to infinity of [n*ln(n)]/[(n+8)^3]

    I've tried direct comparison to k^2/[(n+8)^3], that didnt get me anywhere.
    I've tried a limit comparison with 1/k^2, limit goes to infinity, no good.
    I dont see how the ratio test could work here.

    any pointers? thanks!
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  2. #2
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    Quote Originally Posted by badfishadi View Post
    Hey everyone, I'm doing some practice problems to study for my calculus exam and can't seem to figure this one out. What am I missing?

    sum from n=1 to infinity of [n*ln(n)]/[(n+8)^3]

    I've tried direct comparison to k^2/[(n+8)^3], that didnt get me anywhere.
    I've tried a limit comparison with 1/k^2, limit goes to infinity, no good.
    I dont see how the ratio test could work here.

    any pointers? thanks!
    Try this

     <br />
\sum_{n=1}^{\infty} \frac{n \ln n}{(n+8)^3} \le \sum_{n=1}^{\infty} \frac{n \ln n}{n^3} = \sum_{n=1}^{\infty} \frac{\ln n}{n^2}<br />
then use the integral test on the last series.
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by badfishadi View Post
    Hey everyone, I'm doing some practice problems to study for my calculus exam and can't seem to figure this one out. What am I missing?

    sum from n=1 to infinity of [n*ln(n)]/[(n+8)^3]

    I've tried direct comparison to k^2/[(n+8)^3], that didnt get me anywhere.
    I've tried a limit comparison with 1/k^2, limit goes to infinity, no good.
    I dont see how the ratio test could work here.

    any pointers? thanks!
    I kind of have the same approach as Danny but not as efficient as his.

    Use the limit test

    a_n = \frac{n\ln{n}}{(n+8)^3}

    b_n = \frac{\ln{n}}{n^2}

    so

    \lim_{n \to \infty} \frac{a_n}{b_n} = 1

    so now if b_n converges so does a_n

    Use the integral test to see if b_n converges

    u= \ln{n}

    du = \frac{dn}{n}

    dv = n^{-2}dn

    v = -n^{-1}

    \int^{\infty}_{1} = -\frac{\ln{n}}{n} +\int \frac{dn}{n^2}

    -\frac{\ln{n}}{n} - \frac{1}{n}\bigg|^{\infty}_{1}

    which converges to a number using L' Hopital.

    so since b_n converges so does a_n
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