 # Use the Limit Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{2n^2-1}{3n^5+2n+1} Kyran Hudson 2021-03-08 Answered
Use the Limit Comparison Test to determine the convergence or divergence of the series.
$$\sum_{n=1}^\infty\frac{2n^2-1}{3n^5+2n+1}$$

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself. irwchh

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.