Q. 79 - Calculus

Question

Prove the statements about the convergence or divergence of sequences in Exercises 78–83, referring to theorems in the section as necessary. For each of these statements, assume that r is a real number and p is a positive real number.

${\left(\frac{1}{k}\right)}^p\to \infty$

Step-by-Step Solution

Answer

Hence proved that $\underset{k\to \infty }{\lim }{\left(\frac{1}{k}\right)}^p=0$

Step 1. Given information

The given sequence ${\left(\frac{1}{k}\right)}^p\to \infty$

Step 2. Prove that 1 k p → 0

For $p>0$ , for increasing k, the terms of the sequence $\left\{\frac{1}{k^p}\right\}$ decreases. The sequence $\left\{\frac{1}{k^p}\right\}$ is a decreasing sequence.
It is given that $p>0$, therefore,

$\underset{k\to \infty }{\lim }{\left(\frac{1}{k}\right)}^p=0$

For $\epsilon >0$

$\left|{\left(\frac{1}{k}\right)}^p-0\right|<\epsilon$

Therefore exists an N greater than ${\left(\frac{1}{\epsilon }\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$p$}\right.}$,then

$\left|{\left(\frac{1}{k}\right)}^p-0\right|<\epsilon for k\ge N$

Thus $\left\{\frac{1}{k^p}\right\}$ is a null sequence

Hence, $\underset{k\to \infty }{\lim }{\left(\frac{1}{k}\right)}^p=0$