Kyran Hudson

2021-03-08

Use the Limit Comparison Test to determine the convergence or divergence of the series.
$\sum _{n=1}^{\mathrm{\infty }}\frac{2{n}^{2}-1}{3{n}^{5}+2n+1}$

irwchh

Expert

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
$\sum _{n=1}^{\mathrm{\infty }}\frac{2{n}^{2}-1}{3{n}^{5}+2n+1}=\sum _{n=1}^{\mathrm{\infty }}\frac{{n}^{2}\left(2-\frac{1}{{n}^{2}}\right)}{{n}^{5}\left(3+\frac{2}{{n}^{4}}+\frac{1}{{n}^{5}}\right)}$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{\left(2-\frac{1}{{n}^{2}}\right)}{{n}^{3}\left(3+\frac{2}{{n}^{4}}+\frac{1}{{n}^{5}}\right)}$
So, $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{\left(2-\frac{1}{{n}^{2}}\right)}{{n}^{3}\left(3+\frac{2}{{n}^{4}}+\frac{1}{{n}^{5}}\right)}$
Let another series ${b}_{n}=\frac{1}{{n}^{3}}$
${b}_{n}$ is convergent p-series since p=3
Now,
$\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{\left(2-\frac{1}{{n}^{2}}\right)}{{n}^{3}\left(3+\frac{2}{{n}^{4}}+\frac{1}{{n}^{5}}\right)}}{\frac{1}{{n}^{3}}}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{\left(2-\frac{1}{{n}^{2}}\right)}{\left(3+\frac{2}{{n}^{4}}+\frac{1}{{n}^{5}}\right)}$
$=\frac{2-0}{3+0+0}$
$=\frac{2}{3}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}=\frac{2}{3}$
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.

Jeffrey Jordon

Expert