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Does the series $$\sum^{\infty}_{n=1}\frac{2n^{5}+13n^{3}}{n^{\frac{1}{n}}(n^6-n^2+7)}$$ converge or diverge? Justify.

I know that it diverges. I am trying to use the comparison test to prove it but I am having trouble finding a smaller series that diverges. Any help would be greatly appreciated.

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  • $\begingroup$ Are you forced to use the comparison test? Otherwise the limit comparison test will suffice. $\endgroup$ – Sangchul Lee Mar 10 '18 at 0:13
  • $\begingroup$ I'm not forced to use any particular method, I just thought the comparison test would be the best choice for this. $\endgroup$ – MANONMARS45 Mar 10 '18 at 0:17
  • $\begingroup$ Comparison by inequality are elegant but not so simple to see and justify in general (see DonAntonio solution for this one), with limit comparison test it becomes straightforward en.m.wikipedia.org/wiki/Limit_comparison_test $\endgroup$ – gimusi Mar 10 '18 at 0:17
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HINT

Note that since $n^{\frac{1}{n}}\to 1$

$$\frac{2n^{5}+13n^{3}}{n^{\frac{1}{n}}(n^6-n^2+7)}\sim \frac2n$$

then use limit comparison test with $\sum \frac1n$.

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$$\frac{2n^5+13n^3}{\sqrt[n]n(n^6-n^2+7)}\ge\frac{2n^5}{2n^6}\;\ldots$$

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  • $\begingroup$ Nice bounding (+1) $\endgroup$ – gimusi Mar 10 '18 at 0:20
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    $\begingroup$ Why $\color{red}2\,n^6$ in the denominator? $\endgroup$ – Bernard Mar 10 '18 at 0:27
  • $\begingroup$ I guess since eventually $-n^2+7<0$ Is less than zero and $\sqrt[n] n<2$. Am I wrong? $\endgroup$ – gimusi Mar 10 '18 at 0:59
  • $\begingroup$ @Bernard I think that way it is easier to see that $\;\sqrt[n]n(n^6-n^2+7)\le2n^6\;$ ...WIthout 2 it is also true, of course. $\endgroup$ – DonAntonio Mar 10 '18 at 8:18
  • $\begingroup$ @gimusi: You're right, of course, but I think it had to be detailed in some way. That explains why using equivalence usually makes explanations shorter. $\endgroup$ – Bernard Mar 10 '18 at 9:25
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You can find an equivalent of the term of your series $$ \frac{2n^5+13n^3}{n^{1/n}\left(n^6-n^2+7\right)}\underset{(+\infty)}{\sim}\frac{2n^5}{n^{1/n}n^6}=\frac{2}{n^{1+1/n}}\underset{(+\infty)}{\sim}\frac{2}{n} $$ Then it diverges.

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  • $\begingroup$ “...by limit comparison test” should be added. $\endgroup$ – gimusi Mar 10 '18 at 0:18

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