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Math Help - Comparison & Limit Comparison test for series

  1. #1
    Senior Member mollymcf2009's Avatar
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    Comparison & Limit Comparison test for series

    Can someone see if I am doing this the right way? Or if there is an EASIER way to do this?

    Determine whether the series converges or diverges.

    \sum^{\infty}_{n=1}   \frac{n^2 + 1}{n^2 \sqrt{n}}

    *Pulled out my variables:
    \sum \frac{n^{\frac{4}{2}}}{n^{\frac{5}{2}}}

    = \frac{n}{n^{\frac{1}{2}}} = n^{\frac{1}{2}}

    So,

    \lim_{n \rightarrow \infty} \frac{n^{\frac{4}{2}}}{n^{\frac{5}{2}}} \cdot \frac{1}{n^{\frac{1}{2}}}

    \rightarrow \lim_{n\rightarrow \infty}  \frac{n^2}{n^3}

    \rightarrow \lim_{n\rightarrow \infty}  \frac{1}{n} = 0

    BUT, because n^{\frac{1}{2}} is DIVERGENT then this series is also divergent. Right?

    This stuff is SO abstract! Any help is GREATLY appreciated! Thanks! Molly
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Can someone see if I am doing this the right way? Or if there is an EASIER way to do this?

    Determine whether the series converges or diverges.

    \sum^{\infty}_{n=1} \frac{n^2 + 1}{n^2 \sqrt{n}}

    *Pulled out my variables:
    \sum \frac{n^{\frac{4}{2}}}{n^{\frac{5}{2}}}

    = \frac{n}{n^{\frac{1}{2}}} = n^{\frac{1}{2}}

    So,

    \lim_{n \rightarrow \infty} \frac{n^{\frac{4}{2}}}{n^{\frac{5}{2}}} \cdot \frac{1}{n^{\frac{1}{2}}}

    \rightarrow \lim_{n\rightarrow \infty} \frac{n^2}{n^3}

    \rightarrow \lim_{n\rightarrow \infty} \frac{1}{n} = 0

    BUT, because n^{\frac{1}{2}} is DIVERGENT then this series is also divergent. Right?

    This stuff is SO abstract! Any help is GREATLY appreciated! Thanks! Molly

    you should DIVIDE the terms not multiply
    It's correct to comare your orginal series to \sum^{\infty}_{n=1} \frac{1} {\sqrt{n}}=\infty since p=1/2.

    Then you take the RATIO of the two terms.

    \lim_{n\to\infty}{ {n^2 + 1\over n^2 \sqrt{n}}\over {1\over \sqrt{n}}}=\lim_{n\to\infty} {n^2 + 1\over n^2 }=1


    ANOTHER point, the sequence \frac{1} {\sqrt{n}} converges to zero

    The series \sum^{\infty}_{n=1} \frac{1} {\sqrt{n}} diverges (goes) to infinity.

    Do not confuse a sequence with a sum.
    Last edited by matheagle; March 25th 2009 at 11:52 AM.
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by matheagle View Post
    you should DIVIDE the terms not multiply
    It's correct to comare your orginal series to \sum^{\infty}_{n=1} \frac{1} {\sqrt{n}}=\infty since p=1/2.

    Then you take the RATIO of the two terms.

    \lim_{n\to\infty}{ {n^2 + 1\over n^2 \sqrt{n}}\over {1\over \sqrt{n}}}=\lim_{n\to\infty} {n^2 + 1\over n^2 }=1


    ANOTHER point, the sequence \frac{1} {\sqrt{n}} converges to zero

    The series \sum^{\infty}_{n=1} \frac{1} {\sqrt{n}} diverges (goes) to infinity.

    Do not confuse a sequence with a sum.
    I did divide, just multiplied by reciprocal. But, I didn't get \frac{1}{n^{1/2}} I got just n^{1/2}  and n^{1/2} goes to infinity as n goes to infinity right? Hence diverges, so because it diverges so does the original series?
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  4. #4
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    \frac{n^2 + 1}{n^2 \sqrt{n}} = \frac{n^2 + 1}{n^{\frac{5}{2}} } {\color{red} \ > \ } \frac{n^2}{n^{\frac{5}{2}}} = \frac{1}{n^{\frac{1}{2}}}

    So by the comparison test ...
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  5. #5
    MHF Contributor matheagle's Avatar
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    Nope, you have a couple of error.

    You set a sum equal to a limit, and the alegbra is wrong too.

    You should have  {n^{4/2}\over n^{5/2}} ={1\over n^{1/2}}.
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