# Use the Direct Comparison Test or the Limit Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{1}{sqrt{n^3+2n}}

Series
Use the Direct Comparison Test or the Limit Comparison Test to determine the convergence or divergence of the series.
$$\sum_{n=1}^\infty\frac{1}{\sqrt{n^3+2n}}$$

2020-12-29
Limit comparison test:
Suppose that we have two series $$\sum a_n$$ and $$\sum b_n$$, define,
$$b_n=\frac{1}{n^{\frac{3}{2}}}$$
Where L is positive and finite then either both series are convergent or both series divergent.
Given that,
$$\sum_{n=1}^\infty\frac{1}{\sqrt{n^3+2n}}$$
Since we have,
$$\sum_{n=1}^\infty\frac{1}{\sqrt{n^3+2n}}$$
Here
$$a_n=\frac{1}{\sqrt{n^3+2n}}$$
So,
$$b_n=\frac{1}{n^{\frac{3}{2}}}$$
Now evaluate $$b_n=\frac{1}{n^{\frac{3}{2}}}$$
$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{1}{\sqrt{n^3+2n}}}{\frac{1}{n^{\frac{3}{2}}}}$$
$$=\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^3+2n}}\times n^{\frac{3}{2}}\right)$$
$$=\lim_{n\to\infty}\left(\frac{1}{n^{\frac{3}{2}}\sqrt{1+\frac{2}{n^2}}}\times n^{\frac{3}{2}}\right)$$
$$=\lim_{n\to\infty}\left(\frac{1}{\sqrt{1+\frac{2}{n^2}}}\right)$$
$$=1$$
Since $$\lim_{n\to\infty}\frac{a_n}{b_n}=L=1$$ which is finite and positive.
Since using p-series test [$$\sum\frac{1}{n^p}$$ is convergent if $$p>1$$ or divergent if $$p\leq1$$], $$b_n=\frac{1}{n^{\frac{3}{2}}}$$ is convergent.
Since $$\lim_{n\to\infty}\frac{a_n}{b_n}$$ is finite and positive and $$\sum b_n$$ is convergent so using limit comparison test $$\sum a_n$$ is also convergent.