Chapter 7: Q. 66 (page 605)

Complete the proof of Theorem 7.18 by evaluating the limits
of the sequences.
Prove that ${(\frac{1}{k})}^{p} \to 0$ , when $p > 0$ .

Short Answer

Expert verified

The theorem is hence proved.

Step by step solution

Step 1. Given Information.

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

Step 2. The proof of the theorem.

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

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

It is given that $p > 0$ , therefore, $\lim_{k \to \infty} {(\frac{1}{k})}^{p} = 0$ .

For $ε > 0$ ,

$|{(\frac{1}{k})}^{p} - 0| < ε$

There exists an $N$ greater than ${(\frac{1}{ε})}^{\frac{1}{p}}$ then,

$|{(\frac{1}{k})}^{p} - 0| < ε$ for $k \geq N$

Thus, $\{\frac{1}{k^{p}}\}$ is a null sequence.

Hence, $\lim_{k \to \infty} {(\frac{1}{k})}^{p} = 0$ .

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Complete the proof of Theorem 7.18 by evaluating the limits
of the sequences.
Prove that ${(\frac{1}{k})}^{p} \to 0$ , when $p > 0$ .

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. The proof of the theorem.

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Complete the proof of Theorem 7.18 by evaluating the limitsof the sequences.Prove that 1kp→0, when p>0.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. The proof of the theorem.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

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Applied Mathematics

Theoretical and Mathematical Physics

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Probability and Statistics

Study anywhere. Anytime. Across all devices.

Complete the proof of Theorem 7.18 by evaluating the limits
of the sequences.
Prove that ${(\frac{1}{k})}^{p} \to 0$ , when $p > 0$ .