Section 10.5: Power Series
Definition: A power series centered at x = a is a series of the form
∞
X
cn (x − a)n ,
n=0
where x is a variable the the cn ’s are constants called the coefficients of the series.
Note: The value of x determines the convergence or divergence of a power series. We are
interested in finding all values of x for which a given power series converges.
Theorem: For a given power series
∞
X
cn (x − a)n , there are only three possibilities:
n=0
1. The series converges only for x = a.
2. The series converges for all x.
3. There is a number R > 0 such that the series converges if |x − a| < R and diverges if
|x − a| > R. In this case, the number R is called the radius of convergence of the
power series.
Definition: The set of all values of x for which a given power series converges is called the
interval of convergence of the power series. The radius of convergence is half of the
length of the interval of convergence.
Example: Find the interval and radius of convergence for the given power series.
∞
X
2n xn
(a)
n2 + 1
n=0
By the Ratio Test,
n+1 n+1
2
an+1 2 x
2x(n2 + 1) n
+
1
= lim = |2x|.
lim · n n = lim n→∞ an n→∞ (n + 1)2 + 1
n→∞ (n + 1)2 + 1 2 x
The series converges for |2x| < 1. That is, for
−1 < 2x < 1
1
1
− <x<
.
2
2
1
Thus, the radius of convergence is R = . The Ratio Test is inconclusive if |2x| = 1.
2
1
So, we consider x = ± separately.
2
1
If x = , then
2
∞
∞
∞
X
X
X
1
1
2n xn
=
≤
< ∞.
2+1
2+1
2
n
n
n
n=0
n=0
n=0
1
The series converges for x = by the Direct Comparison Test.
2
1
If x = − , then
2
∞
∞
X
X
2n xn
(−1)n
=
.
2+1
2+1
n
n
n=0
n=0
The series converges for x = −
1
by the Alternating Series Test. Indeed,
2
1
1. lim 2
= 0, and
n→∞ n + 1
∞
1
2.
is clearly decreasing.
n2 + 1 n=1
1 1
The interval of convergence is − , .
2 2
(b)
∞
X
(−2)n (2x − 1)n
n=0
n+1
By the Ratio Test,
(−2)n+1 (2x − 1)n+1
(−2)(2x − 1)(n + 1) n
+
1
= lim = |4x − 2|.
lim ·
n→∞ n+2
(−2)n (2x − 1)n n→∞ n+2
The series converges for |4x − 2| < 1. That is,
−1 < 4x − 2 < 1
1 < 4x < 3
1
3
<x< .
4
4
1
Thus, the radius of convergence is R = .
4
1
If x = , then
4
∞
∞
X
(−2)n (2x − 1)n X 1
=
.
n
+
1
n
+
1
n=0
n=0
1
This series diverges for x =
by Limit Comparison Test with the harmonic series.
4
Indeed,
1
n lim ·
= 1 ∈ (0, ∞).
n + 1 1
3
If x = , then
4
∞
∞
X
(−2)n (2x − 1)n X (−1)n
=
.
n+1
n+1
n=0
n=0
The series converges for x =
3
by the Alternating Series Test. Indeed,
4
1
1. lim
= 0, and
n→∞ n + 1
∞
1
2.
is decreasing.
n + 1 n=0
1 3
The interval of convergence is
, .
4 4
(c)
∞
X
n!(x − 4)n
n=0
9n+1
By the Ratio Test,
n+1
(n + 1)(x − 4) (n + 1)!(x − 4)n+1
9
= lim = ∞.
·
lim n→∞ 9n+2
n!(x − 4)n n→∞ 9
The series converges only for x = 4. So the interval of convergence is the singleton
{x = 4} and the radius of convergence is R = 0.
(d)
∞
X
(−1)n (3x)n
n=0
(2n + 1)!
By the Ratio Test,
(−1)n+1 (3x)n+1 (2n + 1)! −3x
= lim = 0.
lim ·
n→∞
(2n + 3)!
(−1)n (3x)n n→∞ (2n + 3)(2n + 2) The series converges for all x ∈ R. So the interval of convergence is (−∞, ∞) and the
radius of convergence is R = ∞.
(e)
∞
X
3n (x − 1)n
n=2
ln n
By the Ratio Test,
n+1
3 (x − 1)n+1
3(x − 1) ln n ln
n
= 3|x − 1|.
= lim lim · n
n→∞ ln(n + 1)
3 (x − 1)n n→∞ ln(n + 1) The series converges for 3|x − 1| < 1. That is, for
−1 < 3x < 1
1
1
− <x<
.
3
3
1
Thus, the radius of convergence is R = .
3
4
If x = , then
3
∞
∞
∞
X
X
3n (x − 1)n X 1
1
=
≥
.
ln
n
ln
n
n
n=2
n=2
n=2
The series diverges by the Direct Comparison Test.
2
If x = , then
3
∞
∞
X
3n (x − 1)n X (−1)n
=
.
ln
n
ln
n
n=2
n=2
The series converges for x =
2
by the Alternating Series Test. Indeed,
3
1
1. lim
= 0, and
n→∞ ln n
∞
1
2.
is decreasing.
ln n n=2
2 4
The interval of convergence is
,
.
3 3