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For more information, please see full course syllabus of Calculus BC

For more information, please see full course syllabus of Calculus BC

## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Related Books

### Power Series Convergence

- Power series:
- Power series can be convergence on an interval of
*x* - Test for power series convergence:
- Use ratio test to check for absolute convergence
- If not absolutely convergent, solve for
*x*value when it is absolutely convergent - Check
*x*bounds for convergence by plugging in*x*to create series

### Power Series Convergence

Find the values of x such that ∑

_{n = 1}^{∞}[((x − 2)^{n})/n] converges- Apply the Ratio Test
- lim
_{n →∞}| [(a_{n + 1})/(a_{n})] | = lim_{n →∞}| [((x − 2)^{n + 1})/(n + 1)]( [n/((x − 2)^{n})] ) | = lim_{n →∞}| [((x − 2)n)/(n + 1)] |[n/(n + 1)][n/n] = 1

| x − 2 | < 1 − 1 < x − 2 < 11 < x < 3

Find the values of x such that ∑

_{n = 1}^{∞}[((x + 1)^{n})/(n^{2})] converges- Apply the Ratio Test
- lim
_{n →∞}| [(a_{n + 1})/(a_{n})] | = lim_{n →∞}| [((x + 1)^{n + 1})/((n + 1)^{2})]( [(n^{2})/((x + 1)^{n})] ) | = lim_{n →∞}| [((x + 1)n^{2})/(( n + 1 )^{2})] |[(n^{2})/(( n + 1 )^{2})][(n^{2})/(n^{2})] = 1

| x + 1 | < 1 − 1 < x + 1 < 1 − 2 < x < 0

Find a value of x such that ∑

_{n = 1}^{∞}[((3x^{2}− 1)^{n})/(n^{3/2})] converges- Apply the Ratio Test
- lim
_{n →∞}| [(a_{n + 1})/(a_{n})] | = lim_{n →∞}| [((3x^{2}− 1)^{n + 1})/((n + 1)^{3/2})]( [(n^{3/2})/((3x^{2}− 1)^{n})] ) | = lim_{n →∞}| [((3x^{2}− 1)n^{3/2})/(( n + 1 )^{3/2})] |[(n^{3/2})/(( n + 1 )^{3/2})][(n^{3/2})/(n^{3/2})] = 1 - Find the domain of convergence
- | 3x
^{2}− 1 | < 1

0 < 3x

^{2}< 20 < x^{2}< [2/3]0 < x <√{[2/3]}Find the values of x such that ∑

_{n = 1}^{∞}[n/((x − 1)^{n})] converges- Apply the Ratio Test
- lim
_{n →∞}| [(a_{n + 1})/(a_{n})] | = lim_{n →∞}| [(n + 1)/((x − 1)^{n + 1})]( [((x − 1)^{n})/n] ) | = lim_{n →∞}| [(n + 1)/((x − 1)n)] |[(n + 1)/n][n/n] = 1 - Find the domain of convergence
- [1/(| x − 1 |)] < 1
- [1/(x − 1)] < 1x > 2

[1/(x − 1)] >− 1x < 0

Determine if the series ∑

_{n = 1}^{∞}[(x^{n})/(e^{n})] converges absolutely, and if so what's the domain of x- Apply the Ratio Test
- lim
_{n →∞}| [(a_{n + 1})/(a_{n})] | = lim_{n →∞}| [((x)^{n + 1})/(e^{n + 1})]( [(e^{n})/(x^{n})] ) | = lim_{n →∞}| [x/e] |[x/e]

| [x/e] | < 1 − 1 < [x/e] < 1 − e < x < e

Determine if the series ∑

_{n = 1}^{∞}[((x + 1)^{n})/n!] converges absolutelyApply the Ratio Test

lim

lim

_{n →∞}| [(a_{n + 1})/(a_{n})] | = lim_{n →∞}| [((x + 1)^{n + 1})/((n + 1)!)]( [n!/((x + 1)^{n})] ) | = lim_{n →∞}| [(x + 1)/(( n + 1 ))] | = 0 < 1 The series converges, but does not converge absolutelyFind a value of x such that ∑

_{n = 1}^{∞}[(( − 1)^{n})/(n + 3)] converges- Apply the Ratio Test
- lim
_{n →∞}| [(a_{n + 1})/(a_{n})] | = lim_{n →∞}| [(x^{n + 1})/(n + 4)]( [(n + 3)/(x^{n})] ) | = lim_{n →∞}| x[(n + 3)/(n + 4)] ||x|< 1 - Find the domain of convergence
- | x | < 1
- − 1 < x < 1

Test values

Note that if x = − 1, the series converges via Alternate Series Test. Thus the range is actually

− 1 ≤ x < 1

Note that if x = − 1, the series converges via Alternate Series Test. Thus the range is actually

− 1 ≤ x < 1

Find the values of x such that ∑

_{n = 1}^{∞}[(e^{n})/((x − 12)^{n})] converges- Apply the Ratio Test
- lim
_{n →∞}| [(a_{n + 1})/(a_{n})] | = lim_{n →∞}| [(e^{n + 1})/((x − 12)^{n + 1})]( [((x − 12)^{n})/(e^{n})] ) | = lim_{n →∞}| [e/((x − 12))] | < 1| [e/((x − 12))] | < 1 - Find the domain of convergence
- | [e/((x − 12))] | < 1
- [e/((x − 12))] < 1e + 12 < x

[e/((x − 12))] >− 112 − e > x

Determine if the series ∑

_{n = 1}^{∞}[(x^{n})/n!] converges absolutely, and if so what's the domain of xApply the Ratio Test

lim

lim

_{n →∞}| [(a_{n + 1})/(a_{n})] | = lim_{n →∞}| [((x)^{n + 1})/((n + 1)!)]( [n!/(x^{n})] ) | = lim_{n →∞}| [x/(n + 1)] |0 < 1 The series converges, but does not converge absolutelyDetermine if the series ∑

_{n = 1}^{∞}[(( − 1)^{n}x^{n})/(n^{2})] converges absolutely if so what's the domain of x- Apply the Ratio Test
- lim
_{n →∞}| [(a_{n + 1})/(a_{n})] | = lim_{n →∞}| [(x^{n + 1})/((n + 1)^{2})]( [(n^{2})/(x^{n})] ) | = lim_{n → ∞}| [(xn^{2})/(( n + 1 )^{2})] || x | < 1 - Determine convergence
- | x | < 1

− 1 < x < 1

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Power Series Convergence

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Power Series 0:09
- Definition
- Radius & Interval of Convergence 2:07
- Example 1 2:28
- Example 2 4:18
- Example 3 6:20
- Example 4 9:11

0 answers

Post by Jorge Sardinas on June 3, 2014

In the 3 rd example he says that it is x^(n+1) when it is SUPPOSED to be x^(n-1).

1 answer

Last reply by: John Zhu

Tue Aug 13, 2013 8:30 PM

Post by Timothy Davis on August 5, 2013

How did u make the x -1 go away. U mentioned something about "rational rules. This was around 5:15 into your lecture on Power Series example 2. Somehow (x -1) magically just disappeared!

0 answers

Post by Chung Teak Joon on May 1, 2012

Example 1 seems to be wrong to me....because in the previous video, you said (an) convergies if L is smaller than zero.....

I think you did it in the other way, or the ratio test video's content is simply wrong.

1 answer

Last reply by: Yiru Shi

Wed Feb 10, 2016 2:44 PM

Post by robin joshi on April 25, 2012

The video is absolutely wrong here!!

1 answer

Last reply by: John Zhu

Mon Aug 12, 2013 9:26 PM

Post by William Hamilton on April 19, 2012

When L<1, the series DOES converge absolutely. I think you had it backwards here.