Use the Limit Comparison Test to determine the convergence or divergence of the series. ∑n=1∞2n2−13n5+2n+1

We have to check the given series
$\sum _{n=1}^{\mathrm{\infty }}\frac{2n^2-1}{3n^5+2n+1}$ is convergent or divergent using the limit comparision test.
According to limit comparision test if two series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ 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 $\sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^{\mathrm{\infty }}{a}_n\frac{2n^2-1}{3n^5+2n+1}$
Take common highest power n from denominator and numerator we get
$\sum _{n=1}^{\mathrm{\infty }}\frac{2n^2-1}{3n^5+2n+1}=\sum _{n=1}^{\mathrm{\infty }}\frac{n^2(2-\frac{1}{n^2})}{n^5(3+\frac{2}{n^4}+\frac{1}{n^5})}$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{(2-\frac{1}{n^2})}{n^3(3+\frac{2}{n^4}+\frac{1}{n^5})}$
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.

Answer is given below (on video)