Answer
Conveges
Work Step by Step
Th. 9.13 Limit Comparison Test:
If $a_{n} > 0, b_{n} > 0$, and $\displaystyle \lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=L,$
where $L$ is finite and positive, then
$\displaystyle \sum_{n=1}^{\infty}a_{n}$ and $\displaystyle \sum_{n=1}^{\infty}b_{n}$ either both converge or both diverge.
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Let $a_{n}=\displaystyle \frac{2n^{2}-1}{3n^{5}+2n+1}$.
...If the numerator had a leading power of $n^{5} $
the limit would be easily found, which is why we observe
$\displaystyle \sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}\frac{1}{n^{3}}$
which is a convergent p-series $(p > 1)$.
Test the hypothesis of the theorem:
$\displaystyle \lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}$ =$\displaystyle \lim_{n\rightarrow\infty}\frac{\frac{2n^{2}-1}{3n^{5}+2n+1}}{\frac{1}{n^{3}}}=\lim_{n\rightarrow\infty}\frac{2n^{5}-n^{3}}{3n^{5}+2n+1}$
... divide both numerator and denominator with $ n^{5}$...
$=\displaystyle \lim_{n\rightarrow\infty}\frac{2-\frac{1}{n^{2}}}{3+\frac{2}{n^{4}}+\frac{1}{n^{5}}}=\frac{2}{3}=L$ (finite and positive)
So,
$\displaystyle \sum_{n=1}^{\infty}a_{n}$ converges as well (since they both converge).