Question

Use the comparison test to explain why the series $\sum _{k=1}^{\infty }\frac{1}{\sqrt[\alpha ]{k}}$diverges when $\alpha$is an integer greater than $1$

Short answer

The series$\sum _{k=1}^{\infty }\frac{1}{\sqrt[\alpha ]{k}}$diverges when$\alpha$is greater than$1$

Step-by-step solution

Step 1. Given information

An series is given as$\sum _{k-1}^{\infty }\frac{1}{\sqrt[\alpha ]{k}}$

Step 2. Applying comparison test

Terms of the given series are positive.

Now the series $\sum _{k=1}^{\infty }{b}_k$can be written as

$\sum _{k=1}^{\infty }{b}_k=\frac{1}{\sqrt[a]{k}}$

After that the ratio is $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}$can be written as:

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{\frac{1}{\sqrt[a]{k}}}{\frac{1}{\sqrt[a]{k}}} \\ =\underset{i\to \infty }{\lim }1 \\ =1 \end{gathered}$$

The value of $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=1$is a non-zero finite number.

Considering the conditions $a>1\&\frac{1}{a}<1$, the series $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{1/a}}$is divergent by p-series. Therefore the series$\sum _{k=1}^{\infty }{a}_k$ is also divergent