Transcript Convergence: nth-Term Test, Comparing Nonegative Series

Convergence: nth-Term Test,
Comparing Non-negative Series,
Ratio Test
Power Series and Convergence
We have written statements like:
ln 1 + 𝑥 = 𝑥
𝑥2
−
2
𝑥3
+
3
− ⋯ + −1
𝑛
𝑛−1 𝑥
𝑛
+⋯
But we have not talked in depth about what values of 𝑥
make the identity true.
Example: Investigate whether or not 𝑥 = 0.5 makes the
sentence above true? 𝑥 = 100?
.52 .53
+
− ⋯ + −1
𝑥 = 0.5: ln 1 + .5 = .5 −
2
3
𝑛
.5
𝑛−1
+⋯
𝑛
1002 1003
+
− ⋯ + −1
𝑥 = 100: ln 1 + 100 = 100 −
2
3
Appears to
Converge.
𝑛
100
Appears to
𝑛−1
+ ⋯ Diverge.
𝑛
We need better ways to determine the values of 𝒙 that make the series converge.
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1−𝑥
:
1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯ + 𝑥 𝑛−1 + ⋯
Investigate the partial sums of the series and compare the
𝟏
results to 𝒚 =
on a graph.
𝟏−𝒙
𝒚𝟏
𝒚𝟐
𝒚𝟑
𝒚𝟒
𝒚𝟓
𝒚𝟔
𝟏
=
𝟏−𝒙
=𝟏
=𝟏+𝒙
= 𝟏 + 𝒙 + 𝒙𝟐
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑 + 𝒙𝟒
Window: −2 ≤ 𝑥 ≤ 2 and −10 ≤ 𝑦 ≤ 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1−𝑥
:
1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯ + 𝑥 𝑛−1 + ⋯
Investigate the partial sums of the series and compare the
𝟏
results to 𝒚 =
on a graph.
𝟏−𝒙
𝒚𝟏
𝒚𝟐
𝒚𝟑
𝒚𝟒
𝒚𝟓
𝒚𝟔
𝟏
=
𝟏−𝒙
=𝟏
=𝟏+𝒙
= 𝟏 + 𝒙 + 𝒙𝟐
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑 + 𝒙𝟒
Window: −2 ≤ 𝑥 ≤ 2 and −10 ≤ 𝑦 ≤ 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1−𝑥
:
1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯ + 𝑥 𝑛−1 + ⋯
Investigate the partial sums of the series and compare the
𝟏
results to 𝒚 =
on a graph.
𝟏−𝒙
𝒚𝟏
𝒚𝟐
𝒚𝟑
𝒚𝟒
𝒚𝟓
𝒚𝟔
𝟏
=
𝟏−𝒙
=𝟏
=𝟏+𝒙
= 𝟏 + 𝒙 + 𝒙𝟐
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑 + 𝒙𝟒
Window: −2 ≤ 𝑥 ≤ 2 and −10 ≤ 𝑦 ≤ 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1−𝑥
:
1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯ + 𝑥 𝑛−1 + ⋯
Investigate the partial sums of the series and compare the
𝟏
results to 𝒚 =
on a graph.
𝟏−𝒙
𝒚𝟏
𝒚𝟐
𝒚𝟑
𝒚𝟒
𝒚𝟓
𝒚𝟔
𝟏
=
𝟏−𝒙
=𝟏
=𝟏+𝒙
= 𝟏 + 𝒙 + 𝒙𝟐
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑 + 𝒙𝟒
Window: −2 ≤ 𝑥 ≤ 2 and −10 ≤ 𝑦 ≤ 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1−𝑥
:
1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯ + 𝑥 𝑛−1 + ⋯
Investigate the partial sums of the series and compare the
𝟏
results to 𝒚 =
on a graph.
𝟏−𝒙
𝒚𝟏
𝒚𝟐
𝒚𝟑
𝒚𝟒
𝒚𝟓
𝒚𝟔
𝟏
=
𝟏−𝒙
=𝟏
=𝟏+𝒙
= 𝟏 + 𝒙 + 𝒙𝟐
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑 + 𝒙𝟒
Window: −2 ≤ 𝑥 ≤ 2 and −10 ≤ 𝑦 ≤ 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1−𝑥
:
1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯ + 𝑥 𝑛−1 + ⋯
Investigate the partial sums of the series and compare the
𝟏
results to 𝒚 =
on a graph.
𝟏−𝒙
𝒚𝟏
𝒚𝟐
𝒚𝟑
𝒚𝟒
𝒚𝟓
𝒚𝟔
𝟏
=
𝟏−𝒙
=𝟏
=𝟏+𝒙
= 𝟏 + 𝒙 + 𝒙𝟐
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑 + 𝒙𝟒
Window: −2 ≤ 𝑥 ≤ 2 and −10 ≤ 𝑦 ≤ 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1−𝑥
:
1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯ + 𝑥 𝑛−1 + ⋯
𝟏
On (−𝟏, 𝟏), {𝒚𝟐 , 𝒚𝟑 , 𝒚𝟒 , 𝒚𝟓 , 𝒚𝟔 ,…} converges to 𝒚𝟏 =
. The
𝟏−𝒙
polynomial series is a good approximation of 𝒚𝟏 on (−𝟏, 𝟏).
𝒚𝟏
𝒚𝟐
𝒚𝟑
𝒚𝟒
𝒚𝟓
𝒚𝟔
𝟏
=
𝟏−𝒙
=𝟏
=𝟏+𝒙
= 𝟏 + 𝒙 + 𝒙𝟐
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑
= 𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑 + 𝒙𝟒
Window: −2 ≤ 𝑥 ≤ 2 and −10 ≤ 𝑦 ≤ 10
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin 𝑥:
2𝑛+1
𝑥
3
5
𝑥 − 𝑥3! + 𝑥5! + ⋯ + −1 𝑛
2𝑛 + 1 !
Investigate the partial sums of the series and compare the
results to 𝒚 = sin 𝑥 on a graph.
𝒚𝟏 = sin 𝒙
𝒚𝟐 = 𝒙
𝑥3
𝒚𝟑 = 𝑥 − 3!
𝒚𝟒 = 𝑥 −
𝑥3
𝑥5
3! + 5!
𝑥3
𝑥5
𝑥7
𝒚𝟓 = 𝑥 − 3! + 5! − 7!
𝑥3
𝑥5
𝑥7
𝑥9
𝒚𝟔 = 𝑥 − 3! + 5! − 7! + 9!
Window: −7 ≤ 𝑥 ≤ 7 and −5 ≤ 𝑦 ≤ 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin 𝑥:
2𝑛+1
𝑥
3
5
𝑥 − 𝑥3! + 𝑥5! + ⋯ + −1 𝑛
2𝑛 + 1 !
Investigate the partial sums of the series and compare the
results to 𝒚 = sin 𝑥 on a graph.
𝒚𝟏 = sin 𝒙
𝒚𝟐 = 𝒙
𝑥3
𝒚𝟑 = 𝑥 − 3!
𝒚𝟒 = 𝑥 −
𝑥3
𝑥5
3! + 5!
𝑥3
𝑥5
𝑥7
𝒚𝟓 = 𝑥 − 3! + 5! − 7!
𝑥3
𝑥5
𝑥7
𝑥9
𝒚𝟔 = 𝑥 − 3! + 5! − 7! + 9!
Window: −7 ≤ 𝑥 ≤ 7 and −5 ≤ 𝑦 ≤ 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin 𝑥:
2𝑛+1
𝑥
3
5
𝑥 − 𝑥3! + 𝑥5! + ⋯ + −1 𝑛
2𝑛 + 1 !
Investigate the partial sums of the series and compare the
results to 𝒚 = sin 𝑥 on a graph.
𝒚𝟏 = sin 𝒙
𝒚𝟐 = 𝒙
𝑥3
𝒚𝟑 = 𝑥 − 3!
𝒚𝟒 = 𝑥 −
𝑥3
𝑥5
3! + 5!
𝑥3
𝑥5
𝑥7
𝒚𝟓 = 𝑥 − 3! + 5! − 7!
𝑥3
𝑥5
𝑥7
𝑥9
𝒚𝟔 = 𝑥 − 3! + 5! − 7! + 9!
Window: −7 ≤ 𝑥 ≤ 7 and −5 ≤ 𝑦 ≤ 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin 𝑥:
2𝑛+1
𝑥
3
5
𝑥 − 𝑥3! + 𝑥5! + ⋯ + −1 𝑛
2𝑛 + 1 !
Investigate the partial sums of the series and compare the
results to 𝒚 = sin 𝑥 on a graph.
𝒚𝟏 = sin 𝒙
𝒚𝟐 = 𝒙
𝑥3
𝒚𝟑 = 𝑥 − 3!
𝒚𝟒 = 𝑥 −
𝑥3
𝑥5
3! + 5!
𝑥3
𝑥5
𝑥7
𝒚𝟓 = 𝑥 − 3! + 5! − 7!
𝑥3
𝑥5
𝑥7
𝑥9
𝒚𝟔 = 𝑥 − 3! + 5! − 7! + 9!
Window: −7 ≤ 𝑥 ≤ 7 and −5 ≤ 𝑦 ≤ 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin 𝑥:
2𝑛+1
𝑥
3
5
𝑥 − 𝑥3! + 𝑥5! + ⋯ + −1 𝑛
2𝑛 + 1 !
Investigate the partial sums of the series and compare the
results to 𝒚 = sin 𝑥 on a graph.
𝒚𝟏 = sin 𝒙
𝒚𝟐 = 𝒙
𝑥3
𝒚𝟑 = 𝑥 − 3!
𝒚𝟒 = 𝑥 −
𝑥3
𝑥5
3! + 5!
𝑥3
𝑥5
𝑥7
𝒚𝟓 = 𝑥 − 3! + 5! − 7!
𝑥3
𝑥5
𝑥7
𝑥9
𝒚𝟔 = 𝑥 − 3! + 5! − 7! + 9!
Window: −7 ≤ 𝑥 ≤ 7 and −5 ≤ 𝑦 ≤ 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin 𝑥:
2𝑛+1
𝑥
3
5
𝑥 − 𝑥3! + 𝑥5! + ⋯ + −1 𝑛
2𝑛 + 1 !
Investigate the partial sums of the series and compare the
results to 𝒚 = sin 𝑥 on a graph.
𝒚𝟏 = sin 𝒙
𝒚𝟐 = 𝒙
𝑥3
𝒚𝟑 = 𝑥 − 3!
𝒚𝟒 = 𝑥 −
𝑥3
𝑥5
3! + 5!
𝑥3
𝑥5
𝑥7
𝒚𝟓 = 𝑥 − 3! + 5! − 7!
𝑥3
𝑥5
𝑥7
𝑥9
𝒚𝟔 = 𝑥 − 3! + 5! − 7! + 9!
Window: −7 ≤ 𝑥 ≤ 7 and −5 ≤ 𝑦 ≤ 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin 𝑥:
2𝑛+1
𝑥
3
5
𝑥 − 𝑥3! + 𝑥5! + ⋯ + −1 𝑛
2𝑛 + 1 !
On (−∞, ∞), {𝒚𝟐 , 𝒚𝟑 , 𝒚𝟒 , 𝒚𝟓 , 𝒚𝟔 ,…} converges to 𝒚𝟏 = sin 𝒙. The
polynomial series is a good approximation of 𝒚𝟏 on −∞, ∞ .
𝒚𝟏 = sin 𝒙
𝒚𝟐 = 𝒙
𝑥3
𝒚𝟑 = 𝑥 − 3!
𝒚𝟒 = 𝑥 −
𝑥3
𝑥5
3! + 5!
𝑥3
𝑥5
𝑥7
𝒚𝟓 = 𝑥 − 3! + 5! − 7!
𝑥3
𝑥5
𝑥7
𝑥9
𝒚𝟔 = 𝑥 − 3! + 5! − 7! + 9!
Window: −7 ≤ 𝑥 ≤ 7 and −5 ≤ 𝑦 ≤ 5
Proving Sine Converges
Prove the Maclaurin series
∞
𝑘=0
−1
Investigate the error function:
sin 𝑥 for all real 𝑥.
𝑐 𝑥 − 0 𝑛+1
𝑅𝑛 𝑥 =
𝑛+1 !
𝑔 𝑛+1 𝑐 𝑥 − 0 𝑛+1
𝑅𝑛 𝑥 =
𝑛+1 !
𝑔
𝑛+1
𝑐 𝑥 𝑛+1
=
𝑛+1 !
1 ∙ 𝑥 𝑛+1
𝑥 𝑛+1
≤
≤
𝑛+1 !
𝑛+1 !
𝑔
𝑛+1
2𝑛+1
𝑛 𝑥
2𝑛+1 !
converges to
Investigate the limit of the
last statement
𝑥 𝑛+1
=0
lim
𝑛→∞ 𝑛 + 1 !
This means that 𝑅𝑛 𝑥 → 0
for all 𝑥.
Therefore the Maclaurin
series converges for all 𝑥.
When is a Taylor Series a Good
Approximation?
−1 𝑥 2
Consider the Maclaurin Series for 𝑒
0
0
𝑥≠0:
𝑥=0
Investigate the partial sums of the series and compare the
results to 𝒚 =
𝒚𝟏 =
2
−1
𝑥
𝑒
0
𝒚𝟐 = 𝟎
𝒚𝟑 = 𝟎
𝒚𝟒 = 𝟎
𝒚𝟓 = 𝟎
𝒚𝟔 = 𝟎
𝑥≠0
𝑥=0
2
−1
𝑥
𝑒
0
𝑥 ≠ 0 on a graph.
𝑥=0
When is a Taylor Series a Good
Approximation?
−1 𝑥 2
Consider the Maclaurin Series for 𝑒
0
0
𝑥≠0:
𝑥=0
Investigate the partial sums of the series and compare the
results to 𝒚 =
𝒚𝟏 =
2
−1
𝑥
𝑒
0
𝒚𝟐 = 𝟎
𝒚𝟑 = 𝟎
𝒚𝟒 = 𝟎
𝒚𝟓 = 𝟎
𝒚𝟔 = 𝟎
𝑥≠0
𝑥=0
2
−1
𝑥
𝑒
0
𝑥 ≠ 0 on a graph.
𝑥=0
When is a Taylor Series a Good
Approximation?
−1 𝑥 2
Consider the Maclaurin Series for 𝑒
0
0
𝑥≠0:
𝑥=0
Investigate the partial sums of the series and compare the
results to 𝒚 =
𝒚𝟏 =
2
−1
𝑥
𝑒
0
𝒚𝟐 = 𝟎
𝒚𝟑 = 𝟎
𝒚𝟒 = 𝟎
𝒚𝟓 = 𝟎
𝒚𝟔 = 𝟎
𝑥≠0
𝑥=0
2
−1
𝑥
𝑒
0
𝑥 ≠ 0 on a graph.
𝑥=0
When is a Taylor Series a Good
Approximation?
−1 𝑥 2
Consider the Maclaurin Series for 𝑒
0
0
𝑥≠0:
𝑥=0
Investigate the partial sums of the series and compare the
results to 𝒚 =
𝒚𝟏 =
2
−1
𝑥
𝑒
0
𝒚𝟐 = 𝟎
𝒚𝟑 = 𝟎
𝒚𝟒 = 𝟎
𝒚𝟓 = 𝟎
𝒚𝟔 = 𝟎
𝑥≠0
𝑥=0
2
−1
𝑥
𝑒
0
𝑥 ≠ 0 on a graph.
𝑥=0
When is a Taylor Series a Good
Approximation?
−1 𝑥 2
Consider the Maclaurin Series for 𝑒
0
0
𝑥≠0:
𝑥=0
Investigate the partial sums of the series and compare the
results to 𝒚 =
𝒚𝟏 =
2
−1
𝑥
𝑒
0
𝒚𝟐 = 𝟎
𝒚𝟑 = 𝟎
𝒚𝟒 = 𝟎
𝒚𝟓 = 𝟎
𝒚𝟔 = 𝟎
𝑥≠0
𝑥=0
2
−1
𝑥
𝑒
0
𝑥 ≠ 0 on a graph.
𝑥=0
When is a Taylor Series a Good
Approximation?
−1 𝑥 2
Consider the Maclaurin Series for 𝑒
0
0
𝑥≠0:
𝑥=0
Investigate the partial sums of the series and compare the
results to 𝒚 =
𝒚𝟏 =
2
−1
𝑥
𝑒
0
𝒚𝟐 = 𝟎
𝒚𝟑 = 𝟎
𝒚𝟒 = 𝟎
𝒚𝟓 = 𝟎
𝒚𝟔 = 𝟎
𝑥≠0
𝑥=0
2
−1
𝑥
𝑒
0
𝑥 ≠ 0 on a graph.
𝑥=0
When is a Taylor Series a Good
Approximation?
−1 𝑥 2
Consider the Maclaurin Series for 𝑒
0
0
On 𝟎 , {𝒚𝟐 , 𝒚𝟑 , 𝒚𝟒 , 𝒚𝟓 , 𝒚𝟔 ,…} converges to 𝒚𝟏 =
𝑥≠0:
𝑥=0
2
−1
𝑥
𝑒
0
The series is a good approximation of 𝒚𝟏 on 𝟎 .
𝒚𝟏 =
2
−1
𝑥
𝑒
0
𝒚𝟐 = 𝟎
𝒚𝟑 = 𝟎
𝒚𝟒 = 𝟎
𝒚𝟓 = 𝟎
𝒚𝟔 = 𝟎
𝑥≠0
𝑥=0
𝑥 ≠ 0.
𝑥=0
The Convergence Theorem for Power
Series
There are three possibilities for
respect to convergence:
∞
𝑛=0 𝑐𝑛
𝑥−𝑎
𝑛
with
Radius of Convergence
1. There is a positive number 𝑅 such that the series
diverges for 𝑥 − 𝑎 > 𝑅 but converges for 𝑥 − 𝑎 <
𝑅. The series may or may not converge at either of
the endpoints 𝑥 = 𝑎 − 𝑅 and 𝑥 = 𝑎 + 𝑅.
2. The series converges for every 𝑥 (𝑅 = ∞).
3. The series converges at 𝑥 = 𝑎 and diverges
elsewhere (𝑅 = 0).
The set of all values of 𝒙 for which the series converges is the
interval of convergence.
The n-th Term Test
If lim 𝑎𝑛 ≠ 0, then the infinite series
𝑛→∞
∞
𝑛=1 𝑎𝑛
diverges.
OR
When
determining if a
series converges,
always use this
test first!
If the infinite series ∞
𝑛=1 𝑎𝑛 converges,
then lim 𝑎𝑛 = 0.
𝑛→∞
The converse of this statement is NOT true.
If lim 𝑎𝑛 = 0, the infinite series
∞
𝑛=1 𝑎𝑛
does not necessarily converge.
Examples
Use the n-th Term Test to investigate the convergence of
the series.
1.
∞ 2𝑛+1
𝑛=1 𝑛
Since the limit is not 0,
𝟐𝒏 + 𝟏
=𝟐
lim
the series must diverge.
𝒏→∞
𝒏
2.
5
∞
𝑛=1 10𝑛
𝟓
=𝟎
lim
𝒏
𝒏→∞ 𝟏𝟎
Since the limit is 0, the
n-th Term Test is
inconclusive.
Just because the n-th term goes to zero does not mean the
series necessarily converges. We need more tests to
determine if a series converges.
The Finney Procedure for Determining
Convergence
nth-Term Test
Is lim 𝑎𝑛 = 0?
𝑛→∞
Yes or Maybe
?
No
The series
diverges.
Convergent Geometric Series
𝑛−1
The geometric series ∞
𝑎𝑟
converges
𝑛=1
if and only if 𝑟 < 1. If the series converges,
𝑎
its sum is
.
1−𝑟
Where a is the first term and r
is the constant ratio.
Example: Determine if
𝑒 𝑛
∞
(
𝑛=1 𝜋)
converges.
𝒆
𝝅
This is a geometric series with 𝒓 = .
Since
𝒆
𝝅
< 𝟏, the series converges.
This explanation would warrant full credit on the AP Test. Statements like “I
know this is a __ series with __, so it diverges/converges” are acceptable.
The Finney Procedure for Determining
Convergence
nth-Term Test
Is lim 𝑎𝑛 = 0?
The series
diverges.
No
𝑛→∞
Yes or Maybe
Geometric Series Test
Is 𝑎𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + ⋯ ?
No
?
Yes
Converges to
𝑎
if 𝑟 < 1.
1−𝑟
Diverges if
𝑟 > 1.
Example
𝑛! 2
∞
𝑛=0 2𝑛 !
Investigate the series
to see if it
diverges.
Notice the series is NOT geometric.
∞
𝑛=0
𝑛! 2
1 1 1
1
1
=1+ + +
+
+
+⋯
2𝑛 !
2 6 20 70 252
Using partial sums, it appears the series converges:
1; 1.5; 1.667; 1.717; 1.721
Similar to a Geometric Series, we can investigate the ratio between terms.
a1
a0

1 2
a3
 0.5;
1
a2
a1

1 6
1 2
a2
 0.33;

1 20
a5
 0.3;
1 6
a4
a3

1 70
a4

1 252
 0.28
1 70
 0.29;
1 20
Similar to a converging Geometric Series, the limit of the “common” ratio appears to be
less than 1. According to the next theorem. This means the series converges.
The Ratio Test
Suppose the limit
𝑎𝑛+1
lim
𝑛→∞ 𝑎𝑛
= 𝐿 either exists
or is infinite. Then:
1. If 𝐿 < 1, the series
∞
𝑛=1 𝑎𝑛 converges.
Each successive terms are getting smaller.
∞
𝑛=1 𝑎𝑛 diverges.
Each successive terms are getting larger.
2. If 𝐿 > 1, the series
3. If 𝐿 = 1, the test is inconclusive.
This is the “cousin” test to determining if a Convergent Geometric Series
test. It is identical except: (1) If the “common” ratio is 1, then test fails. (2)
The test does not determine the value the series converges to.
Example 1
Investigate the series
𝑛! 2
∞
𝑛=0 2𝑛 !
to see if it diverges.
Similar to a Geometric Series, we can investigate the ratio between terms.
  n  1  !
 2  n 1 !
2
lim
n 
a n 1
 lim
an
 lim
n 
 n !
 2 n !
2
n 
 n 1
2
 2 n  2  2 n  1 
  n  1  !  2 n  !
2
 lim
 n !  2 n  2  !
2
n 
 lim
2
n  2 n 1
2
n  4n 6n2

1
4
Similar to a converging Geometric Series, this “common” ratio is less than 1.
According to the Ratio Test, this means the series converges.
Notice: Unlike the last
example, this series
depends on a value of x.
Example 2
In #3 on Taylor Series Challenges we worked with the series
1
∞
2𝑛 . Find the radius of convergence.
𝑥
−
2
𝑛=1 32𝑛 2𝑛
If the series
1
lim
n 
a n 1
an

3
lim
2  n 1
 2  n  1 
1
n 
3
 lim
n 
2n
 2n 
2n2
 x  2
 2n   x  2 
2n
2n2
3
 2  n  1   x  2 
3
2n
 x  2
2n
2  n 1
converges, the
ratio is less
than 1.
 x2
2
9
 x  2
2
1
9
x2 3
The interval of
convergence is
The limit depends on n. Separate the n’s.
2
2
2 centered at 2 with
n  x  2
 x  2
n
 x  2

 1
a radius
lim 2
 lim
2
n  3
3
9
 n  1 n   n  1
convergence of 3.
Example 3
Find the radius of convergence for the Maclaurin series for
2
𝑓 𝑥 = 𝑒𝑥 .
∞
2
3
𝑛
𝑥
𝑥
𝑥
We know: 𝑒 𝑥 = 1 + 𝑥 + + + ⋯ + + ⋯ =
2! 3!
𝑛!
Thus: 𝑒
𝑥2
4
6
2𝑛
𝑥
𝑥
𝑥
= 1 + 𝑥2 + + + ⋯ +
+⋯=
2! 3!
𝑛!
Investigate
the ratio:
lim
n 
a n 1
an
𝑛=0
∞
x
 lim
n 
𝑥𝑛
𝑛!
𝑛=0
𝑥 2𝑛
𝑛!
2  n 1
 n  1 !
x
2n
 lim
n 
x
2n2
n!
 n  1 ! x
2n
 lim
n 
n!
The series has an infinite radius.
x
2
n 1
 0
Example 4
Find the radius of convergence for the series
  1
a n 1
Investigate
lim
 lim
the ratio: n   a
n 
n
 n  1 1
5
x
n 1
n 1
  1
n 1
5
x
−1 𝑛−1 𝑥 𝑛
∞
.
𝑛=1
𝑛
5
n
 lim
n 
5
5
n
  1 x
n 1
n
  1
n 1
n 1
n
Since the ratio tests uses
n
n 1
5 x
x
x
an absolute value, the
 lim

powers on -1 do not  lim
n 1 n
n  5
n  5
x
5
affect the limit.
If the series
converges, the
ratio is less
than 1.
x
5
1
x 5
The interval of convergence is
centered at 0 with a radius
convergence of 5.
x
n
White Board Challenge
Use the Ratio Test to determine the radius of
∞ 𝑛+1
convergence for 𝑛=1 𝑛 𝑥.
lim
n 
a n 1
an
 n 1 1
 lim
n 
n 1
n 1
n
x
x
 lim
n 
nn 2
n  n 1
 lim
n 
n2
n 1
 1
The Ratio Test is inconclusive because the
limit is 1. We need to use other tests to
determine if the series converges.
The Direct Comparison Test
Suppose 0 ≤ 𝑎𝑛 ≤ 𝑏𝑛 for all 𝑛 ≥ 𝑁.
Both series must have only positive terms.
• If
• If
∞
𝑛=1 𝑏𝑛
∞
𝑛=1 𝑎𝑛
converges, then ∞
𝑛=1 𝑎𝑛 converges.
diverges, then ∞
𝑛=1 𝑏𝑛 diverges.
Example: Prove
For all 𝑛: 0 ≤
2𝑛
𝑥 2𝑛
𝑛! 2
𝑥 2𝑛
∞
𝑛=0 𝑛! 2
≤
𝑥 2𝑛
𝑛!
converges for all real 𝑥.
𝑥 2𝑛
∞
𝑛=0 𝑛! is the Taylor
Series for 𝑓 𝑥 =
∞ 𝑥
Since 𝑛=0
converges for all real 𝒙,
𝑛!
2
𝑥
𝑒 .
𝑥 2𝑛
∞
𝑛=0 𝑛! 2 converges
all real 𝒙 by the Direct Comparison Test.
for
What about Negative Terms?
Does the Direct Comparison Test fail if there are negative terms?
Consider the
sin 𝑥 𝑛
∞
series 𝑛=0
𝑛!
∞
𝑛
sin 3𝜋
2
𝑛=0
𝑛!
if 𝑥 = 3𝜋
:
2
1
1
= 1 − 1 + 12 − 16 + 24
− 120
…
Using partial sums, it appears the series converges:
1; 0; 0.5; 0.333; 0.375; 0.367
But we don’t have a test to prove the series converges.
Similar to the Ratio Test, what happens when we look at the
absolute value of each term:
∞
𝑛
sin 3𝜋
2
1
1
1
= 1 + 1 + 12 + 16 + 24
+ 120
… + 𝑛!
𝑛!
𝑛=0
Using partial sums, this series ALSO appears to converges:
1; 2; 2.5; 2.667; 2.708; 2.717
Since the new series is less than the original, if we can prove the new
series converges, then the original series must converge.
Absolute Convergence Test
If
∞
𝑛=1 𝑎𝑛 converges, then
∞
𝑛=1 𝑎𝑛 converges.
Unlike the Comparison Test, this test does not require the terms
to be positive.
Example: Prove
Investigate
sin 𝑥 𝑛
𝑛!
sin 𝑥 𝑛
∞
𝑛=0
𝑛!
Notice:
sin 𝑥 𝑛
1
≤
𝑛!
𝑛!
converges for all 𝑥.
∞ 1
𝑛=0 𝑛!
lim
n 
a n 1
an

converges by the Ratio Test:
1
lim
n 
 n  1 !
1
 lim
n 
n!
 n  1 !
 lim
n 
1
n 1
n!
sin 𝑥 𝑛
Since
converges for
𝑛!
sin 𝑥 𝑛
∞
converges for all
𝑛=0
𝑛!
all real 𝑥 by the Direct Comparison Test,
real 𝑥 by the Absolute Convergence Test.

0
Definition of Absolute Convergence
If ∞
𝑛=1 𝑎𝑛 converges, then
absolutely.
∞
𝑛=1 𝑎𝑛
converges
Example: The Alternating Harmonic Series
converges but it does not converge absolutely:
∞
(−1)𝑛−1
1
𝑛
= 1 − 12 + 13 − 14 + ⋯ = ln 2
𝑛=1
∞
(−1)𝑛−1
𝑛=1
∞
1
𝑛
1
𝑛
=
𝑛=1
= 1 + 12 + 13 + 14 + ⋯
This is the divergent
Harmonic Series.
Note about Absolute Convergence
If ∞
𝑛=1 𝑎𝑛 converges, then
absolutely.
∞
𝑛=1 𝑎𝑛
converges
𝑎𝑛+1
𝑎𝑛
The Ratio Test uses an absolute value:
.
Thus, every test that converges by the Ratio
Test also converges absolutely:
∞
𝑛−1
4(−0.5)
𝑛=1
∞
𝑛−1
𝑛=1 4(0.5)
converges absolutely because
converges by by the ratio test
The Ratio Test is incredibly strong