We have to check the given series

$\sum _{n=1}^{\mathrm{\infty}}\frac{2{n}^{2}-1}{3{n}^{5}+2n+1}$ is convergent or divergent using the limit comparision test.

According to limit comparision test if two series $\sum _{n=1}^{\mathrm{\infty}}{a}_{n}$ and $\sum _{n=1}^{\mathrm{\infty}}{b}_{n}$ with ${a}_{n}>0,{b}_{n}>0$ for all n.

Then if $\underset{n\to \mathrm{\infty}}{lim}\frac{{a}_{n}}{{b}_{n}}=c$ with 0 then either both series converges or both series divergent.

Let $\sum _{n=1}^{\mathrm{\infty}}{a}_{n}=\sum _{n=1}^{\mathrm{\infty}}{a}_{n}\frac{2{n}^{2}-1}{3{n}^{5}+2n+1}$

Take common highest power n from denominator and numerator we get

$\sum _{n=1}^{\mathrm{\infty}}\frac{2{n}^{2}-1}{3{n}^{5}+2n+1}=\sum _{n=1}^{\mathrm{\infty}}\frac{{n}^{2}(2-\frac{1}{{n}^{2}})}{{n}^{5}(3+\frac{2}{{n}^{4}}+\frac{1}{{n}^{5}})}$

$=\sum _{n=1}^{\mathrm{\infty}}\frac{(2-\frac{1}{{n}^{2}})}{{n}^{3}(3+\frac{2}{{n}^{4}}+\frac{1}{{n}^{5}})}$

So, $\sum _{n=1}^{\mathrm{\infty}}{a}_{n}=\sum _{n=1}^{\mathrm{\infty}}\frac{(2-\frac{1}{{n}^{2}})}{{n}^{3}(3+\frac{2}{{n}^{4}}+\frac{1}{{n}^{5}})}$

Let another series ${b}_{n}=\frac{1}{{n}^{3}}$

${b}_{n}$ is convergent p-series since p=3

Now,

$\underset{n\to \mathrm{\infty}}{lim}\frac{{a}_{n}}{{b}_{n}}=\underset{n\to \mathrm{\infty}}{lim}\frac{\frac{(2-\frac{1}{{n}^{2}})}{{n}^{3}(3+\frac{2}{{n}^{4}}+\frac{1}{{n}^{5}})}}{\frac{1}{{n}^{3}}}$

$=\underset{n\to \mathrm{\infty}}{lim}\frac{(2-\frac{1}{{n}^{2}})}{(3+\frac{2}{{n}^{4}}+\frac{1}{{n}^{5}})}$

$=\frac{2-0}{3+0+0}$

$=\frac{2}{3}$

$=\underset{n\to \mathrm{\infty}}{lim}\frac{{a}_{n}}{{b}_{n}}=\frac{2}{3}$

which is finite and positive.

Therefore we can conclude by limit comparison test series $\sum _{n=1}^{\mathrm{\infty}}{a}_{n}$ will be convergent.

So given series is convergent.