5 Steps to a 5 AP Calculus BC 2019

Series 357

The series

∑∞

n= 1

(−1)n

2 k+ 1

5 k− 1

=−

3

4

+

5

9

−

1

2

.... Note that limn→∞an=nlim→∞

2 n+ 1

5 n− 1

=nlim→∞

2 n+ 1 /n

5 n− 1 /n

=

nlim→∞

2 +

1

n

5 −

1

n

=

2

5

/=0.

Therefore, according to the Divergence Test, the series diverges.

14.5 Power Series

Main Concepts:Power Series, Radius and Interval of Convergence

A power series is a series of the form

∑∞

n= 1

cnxn, wherec 1 ,c 2 ,c 3 ,...are constants, andxis a

variable.

Radius and Interval of Convergence

A power series centered atx=aconverges only forx=a, for all real values ofx, or for all

xin some open interval (a−R,a+R), called the interval of convergence. The radius of

convergence isR. If the series converges on (a−R,a+R), then it diverges ifx<a−Ror

x>a+R: but convergence or divergence must be investigated individually atx=a−R

and atx=a+R.

Example 1

Find the values ofxfor which the series 1+x+

x^2

2!

+

x^3

3!

+···+

xn

n!

+···converges.

Step 1: Use the ratio test. limn→∞

∣∣

∣∣an+^1

an

∣∣

∣∣=lim

n→∞

∣∣

∣∣ x

n+ 1

(n+1)!

·

n!

xn

∣∣

∣∣=lim

n→∞

∣∣

∣∣ x

n+ 1

∣∣

∣∣= 0

Step 2: The series converges absolutely, so

∑∞

n= 1

xn

n!

converges for all realx.

Example 2

Find the interval of convergence for the series

∑∞

n= 1

(x−2)n

n

.

Step 1: Use the ratio test. limn→∞

∣∣

∣∣an+^1

an

∣∣

∣∣=lim

n→∞

∣∣

∣∣(x−2)

n+ 1

n+ 1

·

n

(x−2)n

∣∣

∣∣=lim

n→∞

∣∣

∣∣n(x−2)

n+ 1

∣∣

∣∣=|x− 2 |

Step 2: The series converges absolutely when|x− 2 | < 1,− 1 < x − 2 < 1, or

1 <x<3.

Step 3: Whenx=1, the series becomes

∑∞

n= 1

(−1)n

n

=− 1 +

1

2

−

1

3

+

1

4

−

1

5

···. Since 1>

1

2

>

1

3

>

1

4

>

1

5

>···and limn→∞

1

n

=0, this alternating series converges. When

x=3, the series becomes

∑∞

n= 1

1

n

= 1 +

1

2

+

1

3

+

1

4

+

1

5

···which is a p-series with

p=1, and therefore diverges. Therefore, the interval of convergence is [1, 3).