Question

Complete the proof of Theorem 7.18 by evaluating the limits

of the sequences.

Prove that ${\left(\frac{1}{k}\right)}^p\to 0$, when $p>0$.

Short answer

The theorem is hence proved.

Step-by-step solution

Step 1. Given Information.

The objective is to prove that${\left(\frac{1}{k}\right)}^p\to 0$.

Step 2. The proof of the theorem.

The sequence $\left\{\frac{1}{k}\right\}$is a decreasing sequence as $k$increases, $\frac{1}{k}$decreases.

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 $\epsilon >0$,

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

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