Q. 11 - Calculus

Question

Use the integral test to show that the series $\sum _{k=2}^{\infty }\frac{1}{k\ln k}$ diverges.

Step-by-Step Solution

Answer

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

Step 1. Given information

A series is given as $\sum _{k=2}^{\infty }\frac{1}{k\ln k}$

Step 2. Integral

Here all conditions of integral such as continuous, positive terms, etc. are true for the given series. So we can do integration.

The integral is ${\int }_{x=2}^{\infty }f\left(x\right)dx={\int }_{x=2}^{\infty }\frac{1}{x\left(\ln x\right)}dx$

Now simplify it as,

$$\begin{gathered} {\int }_{x=2}^{\infty }\frac{1}{x(\ln x{)}^2}dx=\underset{i\to \infty }{\lim }{\int }_{x=2}^1\frac{1}{x\left(\ln x\right)}dx \\ =\underset{k\to \infty }{\lim }{\int }_{u=\ln 2}^{\ln k}\frac{1}{u}du\left(\text{ Put }\ln x=u,\Rightarrow \frac{1}{x}dx=du\right) \\ =\underset{k\to \infty }{\lim }[\ln u{]}_{\ln 2}^{\ln k} \\ =\underset{k\to \infty }{\lim }[\ln \left(\ln k\right)-\ln \left(\ln 2\right) \\ =\infty \end{gathered}$$

Hence the series is divergent.