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.

7.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{2}}^{\mathbf{\infty }}\frac{\mathbf{1}}{\mathbf{n}\mathbf{ }\mathbf{lnn}}$

Short answer

The series$\sum _{n=2}^{\infty }\frac{1}{n \ln n}$ 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 $\sum _{n=2}^{\infty }\frac{1}{n \ln n}$.

Use the integral test in the given series as:

$\underset{}{\overset{\infty }{\int }}\frac{1}{n \ln n} dn$

Now, solve the integral is as follows:

Let, $t=\ln n$, so $dt=\frac{1}{n} dn$.

Solve the integral

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

$\underset{}{\overset{\infty }{\int }}\frac{1}{t} dt$

Now, solve the integral is as follows:

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

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