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Math Help - Series converges or diverges

  1. #1
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    Series converges or diverges

    Determine whether the series converges or diverges.



    I got the correct answer but I'm not sure if my method is correct, or if I made so many mistakes that I ended up with the correct answer. If someone could go over this and tell me if my math is perfect or flawed, that would be fantastic.

    First I set up a comparison series. The (7/3) comes from taking the n^7 out of the cubed root.

    \sum_{n=1}^{\infty} \frac{n}{n^\frac{7}{3}} = \sum_{n=1}^{\infty} \frac{1}{n^\frac{4}{3}}

    This new series is a convergent p-series since p = (4/3), and (4/3) > 1.

    So to test if the original series is also convergent, I have to use either the direct comparison test or the limit test. The limit test looks like it would be a pain to do, so I tried the direct comparison test. Unfortunately, the original series is not less than or equal to the new series, so the direct comparison test would not work. So I used the limit test.

    \lim_{n \to \infty} | \frac{a_{n}}{b_{n}} |

    \lim_{n \to \infty} | \frac{n^{\frac{4}{3}}*n+2}{\sqrt[3]{n^{7}+n^{2}}} |

    \lim_{n \to \infty} | \frac{n^{\frac{7}{3}}+2}{\sqrt[3]{n^{7}+n^{2}}} |

    Seeing that the highest power up top is equal to the highest power on bottom (7/3), I used to shortcut that the limit will end up being the ratio of those two coefficients (in this case, 1/1 = 1).

    So b_n is >0, the limit is 1, which is a positive, finite number, which confirms that the original series is convergent.

    I think the radical is throwing me off... I don't know what to do with it. When I try to take the limit do I divide all the terms by n^7, or n^7/3 since it's under the radical? Or is it fine the way I did it?
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  2. #2
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    n \geqslant 2\, \Rightarrow \,\dfrac{{n + 2}}<br />
{{\sqrt[3]{{n^7  + n^2 }}}} \leqslant \dfrac{{2n}}<br />
{{\sqrt[3]{{n^7 }}}} = \dfrac{2}<br />
{{n^{\frac{4}<br />
{3}} }}
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  3. #3
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    If I understand you correctly, you are comparing it to,

    \frac{2n}{\sqrt[3]{{n^7}}}

    , since the addition of the n^2 in the original denominator is negligible, as well as the addition of the 2 in the original numerator. Though I am a bit confused about where the 2n came from in the numerator in the comparison series? Could I just leave it as a 1 and use the p-series test to prove convergence?
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  4. #4
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    If n\ge 2 then n+2\le 2n. Simple?
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