Q. 12 - Calculus

Question

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

Step-by-Step Solution

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 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 $\frac{1}{n}<1 when n>1$.

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