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Questions on Infinite Sequences and Series - Calculus II | MATH 1920, Exams of Calculus

Pellissippi State Technical Community College Calculus

Material Type: Exam; Class: Calculus II; Subject: Mathematics; University: Pellissippi State Technical Community College; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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INFINITE SEQUENCES AND SERIES

8.5 Power Series

Objective: Determine if a power series converges or diverges

I. Power series: =

n

n0

cx

∞

=

∑23

0123

ccxcxcx...

++++

A. A power series may converge for some values of x and diverge for others

B. The sum of the series is a function

1. f(x) =

23n

0123n

ccxcxcx...cx...

++++++

2. Domain is set of all x for which series converges

3. f(x) is similar to a polynomial, but has infinitely many terms

C. If c n = 1 for all n, the power series becomes the geometric series n

n0

1

x

1x

∞

=

=−

∑

1. The series converges if | x | < 1; the interval of convergence is (1,1).−

2. Radius of convergence is 1.

II. =

n

n0

c(xa)

∞

=

−

∑

23n

0123n

cc(xa)c(xa)c(xa)...c(xa)...

+−+−+−++−+

A. Power series centered at a.

B. even if x = a. [for series]

0

(xa)1

−=

C. Always converges if x = a.

III. Example 1: For what values of x is the series convergent?

n

n0

n!x

∞

=

∑

A. Ratio test: If x 0,

≠

n1

n1 n

nnn

n

a(n1)!x

limlimlim(n1)|x|

an!x

+

→∞→∞→∞

+

==+=

n

|x|lim(n1)

→∞

+=∞

B. Thus , the series diverges when x0.≠

C. If x = 0, all terms are zero and the series converges.

D. Interval of convergence is {0} . [A collapsed interval]

E. Radius of convergence R = 0.

IV. Example 2 : For what values of x is the series convergent?

n

n1

(x3)

n

∞

=

−

∑

A. Use the ratio test to show that the series is absolutely convergent [and therefore]

Partial preview of the text

Download Questions on Infinite Sequences and Series - Calculus II | MATH 1920 and more Exams Calculus in PDF only on Docsity!

INFINITE SEQUENCES AND SERIES

8.5 Power Series

Objective: Determine if a power series converges or diverges

I. Power series: =

n n

n 0

c x

∞

=

∑

2 3 c 0 + c x 1 + c x 2 + c x 3 +...

A. A power series may converge for some values of x and diverge for others

B. The sum of the series is a function

1. f(x) =

2 3 n c 0 + c x 1 + c 2 x + c x 3 +... + c xn +...

2. Domain is set of all x for which series converges

3. f(x) is similar to a polynomial, but has infinitely many terms

C. If c n = 1 for all n, the power series becomes the geometric series

n

n 0

x 1 x

∞

=

∑

1. The series converges if | x | < 1; the interval of convergence is ( −1, 1).

2. Radius of convergence is 1.

II. =

n n

n 0

c (x a)

∞

=

∑

2 3 n c 0 + c (x 1 − a) + c 2 (x − a) + c 3 (x − a) +... + c (xn − a) +...

A. Power series centered at a.

B. even if x = a. [for series]

0 (x − a) = 1

C. Always converges if x = a.

III. Example 1: For what values of x is the series convergent?

n

n 0

n!x

∞

=

∑

A. Ratio test: If x ≠ 0,

n 1 n 1

n n n n n

a (^) (n 1)!x lim lim lim (n 1) | x | a (^) n!x

→ ∞ → ∞ → ∞

n

| x | lim (n 1) → ∞

B. Thus , the series diverges when x ≠0.

C. If x = 0, all terms are zero and the series converges.

D. Interval of convergence is {0}. [A collapsed interval]

E. Radius of convergence R = 0.

IV. Example 2 : For what values of x is the series convergent?

n

n 1

(x 3)

n

∞

=

∑

A. Use the ratio test to show that the series is absolutely convergent [and therefore]

convergent when | x – 3 | < 1.

B. − 1 < x − 3 < 1 ⇒ 2 < x < 4 ⇒the open interval of convergence is (2, 4).

C. Radius of convergence R = 1

D. Must test endpoints x = 2 and x = 4 with some other test since ratio test fails at

endpoints.

1. When x = 4, the series become the harmonic series which is divergent.

2. When x = 2, the series converges by the Alternating Series Test.

E. Interval of convergence is [2, 4)

V. Example 3: Bessel function of order 0:

n 2n

(^0) 2n 2

n 0

( 1) x J (x) 2 (n!)

∞

=

A. Use the Ratio Test to show that the series converges for all x.

B. Interval of convergence is ( − ∞, ∞ )⇒ domain of the Bessel function is ( − ∞, ∞).

C. Radius of convergence R = ∞.

VI. For a given power series there are 3 possibilities:

n n

n 0

c (x a)

∞

=

A. The series converges only when x = a. [See Example 1]

B. The series converges for all x. [See Example 3]

C. There is a positive number R such that the series converges if | x − a | <Rand

diverges if | x − a | >R. [See Example 2]

VII. Find radius of convergence and interval of convergence of

n n

n 0

( 3) x

n 1

∞

=

A. The Riot Test can usually be used to find the open interval of convergence.

B. =

n 1 n 1 n 1

n n n^ n n

a (^) ( 3) x n 1 lim lim a (^) n 2 ( 3) x

→ ∞ → ∞

g n

n 3 | x | lim 3 | x | 1 2 n

→ ∞

C. By ratio test: series converges if 3 |x| < 1 and diverges if 3 |x| > 1 ⇒ series

converges if and diverges if

|x| 3

1. Radius of convergence is R =

2. Series converges in but what about the endpoints?

C. If x = , = =

n n

n 0

n 1

∞

=

n 0

n 1

∞

=

Questions on Infinite Sequences and Series - Calculus II | MATH 1920, Exams of Calculus

Related documents

Partial preview of the text

Download Questions on Infinite Sequences and Series - Calculus II | MATH 1920 and more Exams Calculus in PDF only on Docsity!

INFINITE SEQUENCES AND SERIES

8.5 Power Series

Objective: Determine if a power series converges or diverges

I. Power series: =

A. A power series may converge for some values of x and diverge for others

B. The sum of the series is a function

1. f(x) =

2. Domain is set of all x for which series converges

3. f(x) is similar to a polynomial, but has infinitely many terms

C. If c n = 1 for all n, the power series becomes the geometric series

1. The series converges if | x | < 1; the interval of convergence is ( −1, 1).

2. Radius of convergence is 1.

II. =

A. Power series centered at a.

B. even if x = a. [for series]

C. Always converges if x = a.

III. Example 1: For what values of x is the series convergent?

A. Ratio test: If x ≠ 0,

B. Thus , the series diverges when x ≠0.

C. If x = 0, all terms are zero and the series converges.

D. Interval of convergence is {0}. [A collapsed interval]

E. Radius of convergence R = 0.

IV. Example 2 : For what values of x is the series convergent?

A. Use the ratio test to show that the series is absolutely convergent [and therefore]

convergent when | x – 3 | < 1.

B. − 1 < x − 3 < 1 ⇒ 2 < x < 4 ⇒the open interval of convergence is (2, 4).

C. Radius of convergence R = 1

D. Must test endpoints x = 2 and x = 4 with some other test since ratio test fails at

endpoints.

1. When x = 4, the series become the harmonic series which is divergent.

2. When x = 2, the series converges by the Alternating Series Test.

E. Interval of convergence is [2, 4)

V. Example 3: Bessel function of order 0:

A. Use the Ratio Test to show that the series converges for all x.

B. Interval of convergence is ( − ∞, ∞ )⇒ domain of the Bessel function is ( − ∞, ∞).

C. Radius of convergence R = ∞.

VI. For a given power series there are 3 possibilities:

A. The series converges only when x = a. [See Example 1]

B. The series converges for all x. [See Example 3]

C. There is a positive number R such that the series converges if | x − a | <Rand

diverges if | x − a | >R. [See Example 2]

VII. Find radius of convergence and interval of convergence of

A. The Riot Test can usually be used to find the open interval of convergence.

B. =

C. By ratio test: series converges if 3 |x| < 1 and diverges if 3 |x| > 1 ⇒ series

converges if and diverges if

1. Radius of convergence is R =

2. Series converges in but what about the endpoints?

C. If x = , = =