Question

Question: Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals.

13.$\mathbf{\sum }_{\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{n}}^{\mathbf{2}}}{{\mathbf{n}}^{\mathbf{3}}\mathbf{+}\mathbf{1}}$

Short answer

The series$\sum _1^{\infty }\frac{n^2}{n^3+1}$ is divergent.

Step-by-step solution

Definition of convergent and divergent

If the partial sum${\mathbf{\text{S}}}_{\mathbf{n}}$of an infinite series tend to a limit S, then the series is called convergent.

If the partial sum${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series do not approach to a limit, the series is called divergent.

The limiting value S is called the sum of the series.

Apply the integral test

The given series is $\sum _1^{\infty }\frac{n^2}{n^3+1}$.

Use the integral in the given series as:

$\underset{}{\overset{\infty }{\int }}\frac{n^2}{n^3+1} dn$

Now, solve the integral is as follows:

Let, $t=n^3+1$, $dt=3n^2 dn$.

Solve the integral

Substitute the value of t and dt into the integral and change the variable from n into t.

$\underset{}{\overset{\infty }{\int }}\frac{dt}{3t}$

Solve the integral is as follows:

$\begin{array}{rcl}\underset{}{\overset{\infty }{\int }}\frac{dt}{3t}& =& \frac{1}{3}{\left[\ln t\right]}^{\infty }\\ & =& \frac{1}{3}\left[\ln \left(\infty \right)\right]\\ & =& \infty \end{array}$

Here, the series approaches to infinite therefore the given series diverges.