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.

8.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{n}}{{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{4}}$

Short answer

The series $\sum _{n=1}^{\infty }\frac{n}{n^2+4}$ is divergent.

Step-by-step solution

Definition of convergent and divergent

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

If the partial sum${\mathit{S}}_{\mathbf{n}}$ 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 $\sum _{n=1}^{\infty }\frac{n}{n^2+4}$.

Use the integral in the given series as:

$\underset{}{\overset{\infty }{\int }}\frac{n}{n^2+4} dn$

Now, solve the integral is as follows:

Let, $t=n^2+4$, so $dt=2n dn$.

Solve the integral

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

$\underset{}{\overset{\infty }{\int }}\frac{dt}{2t}$

Now, solve the integral is as follows:

$$\begin{gathered} \underset{}{\overset{\infty }{\int }}\frac{dt}{2t}=\frac{1}{2}{\left[\ln t\right]}^{\infty } \\ =\frac{1}{2}\left[\ln \left(\infty \right)\right] \\ =\infty \end{gathered}$$

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