Question

Test for convergence: $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{2}}^{\mathbf{\infty }}\frac{\mathbf{2}{\mathbf{n}}^{\mathbf{3}}}{{\mathbf{n}}^{\mathbf{4}}\mathbf{-}\mathbf{2}}$

Short answer

The limit is greater than 0, so the series$\sum _{n=2}^{\infty }\frac{2n^3}{n^4-2}$ diverges.

Step-by-step solution

Concept and formula used to show that the series diverges

A special comparison test has two parts as follows:

1- The series$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}$ converges if${\mathit{a}}_{\mathbf{n}}\mathbf{⩾}\mathbf{0}$ and $\frac{{\mathbf{a}}_{\mathbf{n}}}{{\mathbf{b}}_{\mathbf{n}}}$tends to a finite limit.

2- The series$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}$ diverges if ${\mathit{a}}_{\mathbf{n}}\mathbf{⩾}\mathbf{0}$and $\frac{{\mathbf{a}}_{\mathbf{n}}}{{\mathbf{b}}_{\mathbf{n}}}$tends to a limit greater than 0.

Calculation to show that the given series diverges

Let$a_n=\frac{2n^3}{n^4-2}$ .

Consider the given series as comparison series as follows:

$\sum _{n=2}^{\infty }\frac{n^3}{n^4}=\sum _{n=2}^{\infty }\frac{1}{n}$

Take

$b_n=\frac{1}{n}$

Here, it is known that$\sum b_n=\sum \frac{1}{n}$ is a divergent series.

Now, use the special comparison test and simplify as follows:

role="math" localid="1664274810995" $$\begin{gathered} li{m}_{n\to \infty }\frac{a_n}{b_n}=li{m}_{n\to \infty }\frac{a_n}{b_n}\left(\frac{\frac{2n^3}{n^4-2}}{\frac{1}{n}}\right) \\ =li{m}_{n\to \infty }\frac{a_n}{b_n}\left(\frac{2n^3}{n^4\left(1-\frac{2}{n^4}\right)}·n\right) \\ =\frac{2}{1-\frac{2}{n^4}} \\ =2 \end{gathered}$$

Here, the $li{m}_{n\to \infty }\frac{a_n}{b_n}=2>0$. So, the given series diverges.