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Math Help - Series Evaluation

  1. #1
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    Series Evaluation

    I need to evaluate this for convergence/divergence:

    \sum_{n=1}^\infty \frac{n+5}{(n^7+n^2)^{1/3}}

    I tried a Limit Comparison Test with \frac{1}{n^{7/3}} in the denominator, but that didn't get me anywhere. I don't see another function to compare this to.

    Can anybody suggest what I might try?

    Thanks.
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  2. #2
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    Re: Series Evaluation

    Quote Originally Posted by joatmon View Post
    I need to evaluate this for convergence/divergence:

    \sum_{n=1}^\infty \frac{n+5}{(n^7+n^2)^{1/3}}

    I tried a Limit Comparison Test with \frac{1}{n^{7/3}} in the denominator, but that didn't get me anywhere. I don't see another function to compare this to.

    Can anybody suggest what I might try?

    Thanks.
    Maybe the integral test - the integral can be evaluated numerically...
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  3. #3
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    Re: Series Evaluation

    Oh boy. That's not going to be a fun one to evaluate. Thanks!
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  4. #4
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    Re: Series Evaluation

    Ever try some algebra?

    \frac{n+5}{(n^{7}+n^{2})^{\frac{1}{3}}} = \frac{n+5}{n\cdot(n^{4}+\frac{1}{n})^{\frac{1}{3}}  }=\frac{1+\frac{5}{n}}{(n^{4}+\frac{1}{n})^{\frac{  1}{3}}}<\frac{1+\frac{5}{n}}{n^{\frac{4}{3}}}

    It can get ugly, but it often leads somewhere useful.
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  5. #5
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    Re: Series Evaluation

    Thanks!
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  6. #6
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    Re: Series Evaluation

    Hello, joatmon!

    I need to evaluate this for convergence/divergence:

    . . \sum_{n=1}^{\infty} \frac{n+5}{(n^7+n^2)^{1/3}}

    For n > 5, we have: . n+5 \:<\: 2n .[1]

    We have: . n^7+n^2 \:>\:n^7 \quad\Rightarrow\quad (n^7+n^2)^{\frac{1}{3}} \:>\:(n^7)^{\frac{1}{3}}

    . . Then: . \frac{1}{(n^7 + n^2)^{\frac{1}{3}}} \:<\:\frac{1}{n^{\frac{7}{3}}} .[2]

    Multiply [1] and [2]: . (n+5)\cdot\frac{1}{(n^7 + n^2)^{\frac{1}{3}}} \;<\; 2n\cdot\frac{1}{n^{\frac{7}{3}}}

    . . . . . . . . We have: . . . . . . . \frac{n+5}{(n^7+n^2)^{\frac{1}{3}}} \;<\;\frac{2}{n^{\frac{4}{3}}}

    Hence: . \sum^{\infty}_{n=1}\frac{n+5}{(n^7+n^2)^{\frac{1}{  3}}} \;<\;\sum^{\infty}_{n=1}\frac{2}{n^{\frac{4}{3}}}


    The given series is less than a convergent p-series.

    . . Therefore, it converges.

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  7. #7
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    Re: Series Evaluation

    Well done. Thanks!

    I was thinking about this a little more and I think that TKHunny's solution doesn't quite work because the final comparative function is not quite a p-series, so it doesn't definitively establish the original function as convergent. Then I saw Soroban's effort, and now I see how to get there.

    Thanks to both of you!
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  8. #8
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    Re: Series Evaluation

    Quote Originally Posted by joatmon View Post
    Well done. Thanks!

    I was thinking about this a little more and I think that TKHunny's solution doesn't quite work because the final comparative function is not quite a p-series, so it doesn't definitively establish the original function as convergent. Then I saw Soroban's effort, and now I see how to get there.

    Thanks to both of you!
    No, you have come to the wrong conclusion. Mine doesn't quite work because I didn't do ALL the work for you. I left you some room to explore and learn on your own, through your personal effort.
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