Q. 38 - Calculus

Question

In Exercises 35–40 use the root test to analyze whether the given series converges or diverges. If the root test is inconclusive, use a different test to analyze the series.

$\sum _{k=2}^{\infty }{\left(\frac{1}{\ln k}\right)}^k$

Step-by-Step Solution

Answer

The given series diverges.

Step 1. Given Information.

The given series is $\sum _{k=2}^{\infty }{\left(\frac{1}{\ln k}\right)}^k.$

Step 2. Determine whether the given series converges or diverges.

By using the root test, let the general term is $a_k={\left(\frac{1}{\ln k}\right)}^k.$

So,

$$\begin{gathered} \rho =\underset{k\to \infty }{\lim }{\left(a_k\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.} \\ \rho =\underset{k\to \infty }{\lim }{\left({\left(\frac{1}{\ln k}\right)}^k\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.} \\ \rho =\underset{k\to \infty }{\lim }\left(\frac{1}{\ln k}\right) \\ \rho =1 \end{gathered}$$

Since it is 1, thus the root test is inconclusive.

Step 3. Using the different test to analyze the series.

We will use the divergence test to analyze the series. The divergence test states that if $\underset{k\to \infty }{\lim }{a}_k$ does not exist or if $\underset{k\to \infty }{\lim }{a}_k\ne 0,$ then the series $\sum _{k=1}^{\infty }{a}_k$ is divergent.

So, let the sequence is $a_k={\left(\frac{1}{\ln k}\right)}^k.$

Now, $\underset{k\to \infty }{\lim }{a}_k\ne 0.$

Hence the series diverges.