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.

Question

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.

Answer 1

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 $\sum \frac1n$.

Answer 2

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.

Answer 3

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