Question

Explain why, if n is an integer greater than 1, the series $\sum _{k=1}^{\infty }\frac{1}{\sqrt[n]{k}}$ diverges.

Short answer

To make the series$\sum _{k=1}^{\infty }\frac{1}{\sqrt[n]{k}}$ divergent, the value of n should be greater than $1$.

Step-by-step solution

Step 1. Given Information.

The series:

$\sum _{k=1}^{\infty }\frac{1}{\sqrt{k}}$

Step 2. p-test series.

The p-test states that:

(i) For $p>1$, the series $\sum _{k=1}^{\infty }\frac{1}{k^p}$converges.

(ii) For $p=1$, the harmonic series $\sum _{k=1}^{\infty }\frac{1}{k}$diverges.

(iii) For $p<1$, the series $\sum _{k=1}^{\infty }\frac{1}{k^p}$diverges.

Step 3. Approximate the series.

$$\begin{gathered} \sum _{k=1}^{\infty }\frac{1}{\sqrt[n]{k}}=\sum _{k=1}^{\infty }\frac{1}{k^{{\displaystyle \frac{1}{n}}}} \\ p=\frac{1}{2} \end{gathered}$$

The value of $\frac{1}{n}<1whenn>1$.

To make the series divergent, the value of n should be greater than $1$.