Chapter 7: Q. 28 (page 631)

In Exercises $21 - 30$ use one of the comparison tests to determine whether the series converges or diverges. Explain how the given series satisfies the hypotheses of the test you use.
$\sum_{k = 1}^{\infty} \frac{1}{{(k - π)}^{2}}$ .

Short Answer

Expert verified

The series $\sum_{k = 1}^{\infty} \frac{1}{{(k - π)}^{2}}$ is convergent.

Step by step solution

Step 1. Given information

$\sum_{k = 1}^{\infty} \frac{1}{{(k - π)}^{2}}$ .

Step 2. The comparison test states that for ∑k=1∞ ak and ∑k=1∞ bk be two series with positive term then,

If $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = L$ , where $L$ is any positive real number, then either both converge or both diverge.
If $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = 0$ , and $\sum_{k = 1}^{\infty} b_{k}$ converges, then $\sum_{k = 1}^{\infty} a_{k}$ converges.
If $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = \infty$ , and " width="9">∑k=1∞bkdiverges, then $\sum_{k = 1}^{\infty} a_{k}$ diverges.

Step 3. The terms of the series ∑k=1∞ 1k-π2 are positive.

Find $\sum_{k = 1}^{\infty} b_{k}$ for the given series.

$\sum_{k = 1}^{\infty} b_{k} = \sum_{k = 1}^{\infty} \frac{1}{k^{2}}$

Next find $\lim_{k \to \infty} \frac{a_{k}}{b_{k}}$ for the given series.

$\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = \lim_{k \to \infty} \frac{\frac{1}{{(k - π)}^{2}}}{\frac{1}{k^{2}}} = \lim_{k \to \infty} \frac{k^{2}}{{(k - π)}^{2}} = \lim_{k \to \infty} \frac{k^{2}}{k^{2} {(1 - \frac{π}{k})}^{2}} = \lim_{k \to \infty} \frac{1}{{(1 - \frac{π}{k})}^{2}} = 1$

Step 4. From the obtained values,

The value of $\lim_{k \to \infty} \frac{a_{k}}{b_{k}} = 1$ which is a finite positive number.

The value of localid="1649179701987" $\sum_{k = 1}^{\infty} b_{k} = \sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ is convergent by $p -$ series test.

Therefore, the value of $\sum_{k = 1}^{\infty} b_{k}$ convergent.

Hence., the given series is convergent.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

In Exercises $21 - 30$ use one of the comparison tests to determine whether the series converges or diverges. Explain how the given series satisfies the hypotheses of the test you use.
$\sum_{k = 1}^{\infty} \frac{1}{{(k - π)}^{2}}$ .

Short Answer

Step by step solution

Step 1. Given information

Step 2. The comparison test states that for ∑k=1∞ ak and ∑k=1∞ bk be two series with positive term then,

Step 3. The terms of the series ∑k=1∞ 1k-π2 are positive.

Step 4. From the obtained values,

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Geometry

Mechanics Maths

Statistics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

In Exercises 21-30use one of the comparison tests to determine whether the series converges or diverges. Explain how the given series satisfies the hypotheses of the test you use.∑k=1∞1k-π2.

Short Answer

Step by step solution

Step 1. Given information

Step 2. The comparison test states that for ∑k=1∞ ak and ∑k=1∞ bk be two series with positive term then,

Step 3. The terms of the series ∑k=1∞ 1k-π2 are positive.

Step 4. From the obtained values,

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Geometry

Mechanics Maths

Statistics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

In Exercises $21 - 30$ use one of the comparison tests to determine whether the series converges or diverges. Explain how the given series satisfies the hypotheses of the test you use.
$\sum_{k = 1}^{\infty} \frac{1}{{(k - π)}^{2}}$ .