Q 44. Find the interval of convergence... [FREE SOLUTION]

91影视

Chapter 8: Q 44. (page 670)

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} {(3 x + 7)}^{2 k + 1}$

Short Answer

Expert verified

The interval of convergence for power series is $R$ .

Step by step solution

Step 1. Given information.

The given power series is ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} {(3 x + 7)}^{2 k + 1}$ .

Step 3. Find the interval of convergence.

Let us assume $b_{k} = \frac{{(- 1)}^{k}}{(2 k + 1)!} {(3 x + 7)}^{2 k + 1}$ and $b_{k + 1} = \frac{{(- 1)}^{k + 1}}{[2 (k + 1) + 1]!} {(3 x + 7)}^{2 (k + 1) + 1}$

Ratio for the absolute convergence is

role="math" localid="1649425467628" $\lim_{k 鈫� 鈭�} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k 鈫� 鈭�} |\frac{\frac{{(- 1)}^{k + 1}}{[2 (k + 1) + 1]!} {(3 x + 7)}^{2 k + 3}}{\frac{{(- 1)}^{k}}{[2 k + 1]!} {(3 x + 7)}^{2 k + 1}}| = \lim_{k 鈫� 鈭�} {|3 x + 7|}^{2} \frac{- 1}{(2 k + 3) (2 k + 2)}$

Now, we evaluate the limit at $k 鈫� 鈭�$ .

So, $\lim_{k 鈫� 鈭�} {|3 x + 7|}^{2} \frac{- 1}{(2 k + 3) (2 k + 2)} = 0$ , that is the value of limit will be zero no matter what value the variable $x$ takes.

Hence by the ratio test the series converges absolutely for every value of $x$ .

Therefore, the interval of convergence of the power series is $R$ .

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} {(3 x + 7)}^{2 k + 1}$

Short Answer

Step by step solution

Step 1. Given information.

Step 3. Find the interval of convergence.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Statistics

Mechanics Maths

Discrete Mathematics

Probability and Statistics

Calculus

Study anywhere. Anytime. Across all devices.

91影视

Find the interval of convergence for power series:鈭�k=0鈭�-1k2k+1!3x+72k+1

Short Answer

Step by step solution

Step 1. Given information.

Step 3. Find the interval of convergence.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Statistics

Mechanics Maths

Discrete Mathematics

Probability and Statistics

Calculus

Study anywhere. Anytime. Across all devices.

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{(2 k + 1)!} {(3 x + 7)}^{2 k + 1}$