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Chapter 7: Q. 78 (page 605)

Prove the statements about the convergence or divergence of sequences in Exercises 78鈥�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.
$k^{p} 鈫� 鈭�$

Short Answer

Expert verified

The given sequence $\{a_{k}\} = \{k^{p}\}$ is divergence

Step by step solution

01

Step 1. Given information

The given sequence $k^{p} 鈫� 鈭�$

02

Step 2. Find the given sequence is increasing or decreasing

The general term of the sequence $\{a_{k}\} = \{k^{p}\} i s a_{k} = k^{p}$

The ratio $\frac{a_{k + 1}}{a_{k}}$ gives

$\frac{a_{k + 1}}{a_{k}} = \frac{{(k + 1)}^{p}}{k^{p}} (s u b s t i t u t i o n} = {(\frac{K + 1}{k})}^{p} = {(1 + \frac{1}{k})}^{p} (s i m p l i f y) > 1 (F o r k > 0)$

Thus, $a_{k + 1} > a_{k}$

The sequence $\{a_{k}\} = \{k^{p}\}$ is strickly increasing sequence

03

Step 3. Find the given sequence is converges or divergent

The sequence $\{a_{k}\} = \{k^{p}\}$ is bounded below as for $p > 0$

$0 < a_{k}$

The sequence $\{a_{k}\} = \{k^{p}\}$ is increasing and there is no upper bound

The monitoring increasing sequence which is bounded above is convergent

The sequence role="math" localid="1649300083524" $\{a_{k}\} = \{k^{p}\}$ is strickly increasing but it is not bounded above.Hence the sequence is divergent

The sequence $\{a_{k}\} = \{k^{p}\}$ is divergent.Therefore,

$\lim_{k 鈫� 鈭�} a_{k} = \lim_{x 鈫� 鈭�} k^{p} = 鈭�$

Hence,for $p > 0$ it is proved that $k^{p} 鈫� 鈭�$

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Most popular questions from this chapter

Explain why a function a(x) has to be continuous in order for us to use the integral test to analyze a series ${鈭�}_{k = 1}^{鈭�} a_{k}$ for convergence.

Improper Integrals: Determine whether the following improper integrals converge or diverge.

${鈭�}_{1}^{鈭�} \frac{1}{x^{2}} d x .$

Determine whether the series ${鈭�}_{k = 0}^{鈭�} \frac{{(- 3)}^{k + 1}}{4^{k - 2}}$ converges or diverges. Give the sum of the convergent series.

Let 0 < p < 1. Evaluate the limit $\lim_{k 鈫� 鈭�} \frac{1 / k \ln k}{1 / k^{p}}$

Explain why we cannot use a p-series with 0 < p < 1 in a limit comparison test to verify the divergence of the series ${鈭�}_{k = 2}^{鈭�} \frac{1}{k \log k}$

Given that $a_{0} = - 3, a_{1} = 5, a_{2} = - 4, a_{3} = 2$ and ${鈭� a_{k}}_{k = 2}^{鈭�} = 7$ , find the value of ${鈭� a_{k}}_{k = 0}^{鈭�}$ .

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