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).