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Thread: Series test.

  1. #1
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    Series test.

    Ok so I am a little stuck on this question, I need to test this series using only the standard comparison test.
    $\displaystyle \sum_{n = 2}^\infty \frac{\left\vert{\sin n } \right\vert{} }{{n}^{2 }-1}$

    I can't compare it to $\displaystyle \frac{1}{{n}^{ 2}}$ can I?
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  2. #2
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    Since $\displaystyle \displaystyle |\sin{n}| \leq 1$, you can compare this series to $\displaystyle \displaystyle \sum_{n = 2}^{\infty}{\frac{1}{n^2 - 1}}$. If you use partial fractions, you should find that this series is telescopic...
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    Thanks, also what about
    $\displaystyle \sum_{n = 0}^\infty \frac{{n}^{3 } }{{n}^{ 4}+1 } $
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  4. #4
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    I would use the integral test, because I can see that $\displaystyle \displaystyle \int{\frac{x^3}{x^4 + 1}\,dx}$ is easily evaluated.

    $\displaystyle \displaystyle \begin{align*} \sum_{n = 0}^{\infty}{\frac{n^3}{n^4 + 1}} &\geq \int_0^{\infty}{\frac{x^3}{x^4 + 1}\,dx} \\ &= \frac{1}{4}\int_0^{\infty}{\frac{4x^3}{x^4 + 1}\,dx} \\ &= \frac{1}{4}\int_1^{\infty}{\frac{1}{u}\,du} \\ &= \frac{1}{4}\left[\lim_{\epsilon \to \infty}\left(\ln{\epsilon}\right) - \ln{1}\right] \\ & \to \infty\end{align*}$.

    So the series is divergent.
    Last edited by Prove It; May 20th 2011 at 10:44 PM.
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  5. #5
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    We can't unfortunately. Only the comparison test and the arithmetic, geometric, harmonic series and $\displaystyle \frac{1}{ {n}^{k } } $ series.

    I let n=$\displaystyle {2}^{ m} $ and got $\displaystyle \frac{1}{{2}^{ m}+\frac{1}{ {2}^{ 3m} }} < \frac{1}{{2}^{ m} } $

    which is geometric and converges. But I'm not at all confident with that.
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  6. #6
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    Ok that gives me something to go off I will try and figure out to see if I can show it is divergent somehow
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  7. #7
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    Ok I think I got it.

    $\displaystyle \sum_{n = 1}^\infty \frac{{n}^{3 } }{ {n}^{4 }+1 } \geqslant \sum_{n = 1}^\infty \frac{1}{n+1}$ which diverges so


    $\displaystyle \sum_{n = 0}^\infty \frac{{n}^{3 } }{ {n}^{4 }+1 }$ must diverge.

    Does this sound right?
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  8. #8
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    Ho do you know that $\displaystyle \displaystyle \frac{n^3}{n^4 + 1} \geq \frac{1}{n + 1}$?
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  9. #9
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    For $\displaystyle n\geqslant 1 \frac{{n}^{3 } }{{n}^{4 } +1 } \geqslant \frac{{n}^{3 } }{{n}^{4 } +{n}^{ 3} } = \frac{1}{ n+1}$
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  10. #10
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    OK I agree with your comparison then
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  11. #11
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    Thanks a lot you were a real help
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