We have to check the given series

\(\sum_{n=1}^\infty\frac{2n^2-1}{3n^5+2n+1}\) is convergent or divergent using the limit comparision test.

According to limit comparision test if two series \(\sum_{n=1}^\infty a_n\) and \(\sum_{n=1}^\infty b_n\) with \(a_n>0,b_n>0\) for all n.

Then if \(\lim_{n\to\infty}\frac{a_n}{b_n}=c\) with 0 then either both series converges or both series divergent.

Let \(\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty a_n\frac{2n^2-1}{3n^5+2n+1}\)

Take common highest power n from denominator and numerator we get

\(\sum_{n=1}^\infty\frac{2n^2-1}{3n^5+2n+1}=\sum_{n=1}^\infty\frac{n^2(2-\frac{1}{n^2})}{n^5(3+\frac{2}{n^4}+\frac{1}{n^5})}\)

\(=\sum_{n=1}^\infty\frac{(2-\frac{1}{n^2})}{n^3(3+\frac{2}{n^4}+\frac{1}{n^5})}\)

So, \(\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{(2-\frac{1}{n^2})}{n^3(3+\frac{2}{n^4}+\frac{1}{n^5})}\)

Let another series \(b_n=\frac{1}{n^3}\)

\(b_n\) is convergent p-series since p=3

Now,

\(\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{(2-\frac{1}{n^2})}{n^3(3+\frac{2}{n^4}+\frac{1}{n^5})}}{\frac{1}{n^3}}\)

\(=\lim_{n\to\infty}\frac{(2-\frac{1}{n^2})}{(3+\frac{2}{n^4}+\frac{1}{n^5})}\)

\(=\frac{2-0}{3+0+0}\)

\(=\frac23\)

\(=\lim_{n\to\infty}\frac{a_n}{b_n}=\frac{2}{3}\)

which is finite and positive.

Therefore we can conclude by limit comparison test series \(\sum_{n=1}^\infty a_n\) will be convergent.

So given series is convergent.