Convergence: nth-Term Test, Comparing Nonegative Series
Download
Report
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 + ๐ฅ = ๐ฅ
๐ฅ2
โ
2
๐ฅ3
+
3
โ โฏ + โ1
๐
๐โ1 ๐ฅ
๐
+โฏ
But we have not talked in depth about what values of ๐ฅ
make the identity true.
Example: Investigate whether or not ๐ฅ = 0.5 makes the
sentence above true? ๐ฅ = 100?
.52 .53
+
โ โฏ + โ1
๐ฅ = 0.5: ln 1 + .5 = .5 โ
2
3
๐
.5
๐โ1
+โฏ
๐
1002 1003
+
โ โฏ + โ1
๐ฅ = 100: ln 1 + 100 = 100 โ
2
3
Appears to
Converge.
๐
100
Appears to
๐โ1
+ โฏ Diverge.
๐
We need better ways to determine the values of ๐ that make the series converge.
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1โ๐ฅ
:
1 + ๐ฅ + ๐ฅ 2 + ๐ฅ 3 + โฏ + ๐ฅ ๐โ1 + โฏ
Investigate the partial sums of the series and compare the
๐
results to ๐ =
on a graph.
๐โ๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐
=
๐โ๐
=๐
=๐+๐
= ๐ + ๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐ + ๐๐
Window: โ2 โค ๐ฅ โค 2 and โ10 โค ๐ฆ โค 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1โ๐ฅ
:
1 + ๐ฅ + ๐ฅ 2 + ๐ฅ 3 + โฏ + ๐ฅ ๐โ1 + โฏ
Investigate the partial sums of the series and compare the
๐
results to ๐ =
on a graph.
๐โ๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐
=
๐โ๐
=๐
=๐+๐
= ๐ + ๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐ + ๐๐
Window: โ2 โค ๐ฅ โค 2 and โ10 โค ๐ฆ โค 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1โ๐ฅ
:
1 + ๐ฅ + ๐ฅ 2 + ๐ฅ 3 + โฏ + ๐ฅ ๐โ1 + โฏ
Investigate the partial sums of the series and compare the
๐
results to ๐ =
on a graph.
๐โ๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐
=
๐โ๐
=๐
=๐+๐
= ๐ + ๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐ + ๐๐
Window: โ2 โค ๐ฅ โค 2 and โ10 โค ๐ฆ โค 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1โ๐ฅ
:
1 + ๐ฅ + ๐ฅ 2 + ๐ฅ 3 + โฏ + ๐ฅ ๐โ1 + โฏ
Investigate the partial sums of the series and compare the
๐
results to ๐ =
on a graph.
๐โ๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐
=
๐โ๐
=๐
=๐+๐
= ๐ + ๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐ + ๐๐
Window: โ2 โค ๐ฅ โค 2 and โ10 โค ๐ฆ โค 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1โ๐ฅ
:
1 + ๐ฅ + ๐ฅ 2 + ๐ฅ 3 + โฏ + ๐ฅ ๐โ1 + โฏ
Investigate the partial sums of the series and compare the
๐
results to ๐ =
on a graph.
๐โ๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐
=
๐โ๐
=๐
=๐+๐
= ๐ + ๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐ + ๐๐
Window: โ2 โค ๐ฅ โค 2 and โ10 โค ๐ฆ โค 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1โ๐ฅ
:
1 + ๐ฅ + ๐ฅ 2 + ๐ฅ 3 + โฏ + ๐ฅ ๐โ1 + โฏ
Investigate the partial sums of the series and compare the
๐
results to ๐ =
on a graph.
๐โ๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐
=
๐โ๐
=๐
=๐+๐
= ๐ + ๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐ + ๐๐
Window: โ2 โค ๐ฅ โค 2 and โ10 โค ๐ฆ โค 10
When is a Taylor Series a Good
Approximation?
1
Consider the Maclaurin Series for 1โ๐ฅ
:
1 + ๐ฅ + ๐ฅ 2 + ๐ฅ 3 + โฏ + ๐ฅ ๐โ1 + โฏ
๐
On (โ๐, ๐), {๐๐ , ๐๐ , ๐๐ , ๐๐ , ๐๐ ,โฆ} converges to ๐๐ =
. The
๐โ๐
polynomial series is a good approximation of ๐๐ on (โ๐, ๐).
๐๐
๐๐
๐๐
๐๐
๐๐
๐๐
๐
=
๐โ๐
=๐
=๐+๐
= ๐ + ๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐
= ๐ + ๐ + ๐๐ + ๐๐ + ๐๐
Window: โ2 โค ๐ฅ โค 2 and โ10 โค ๐ฆ โค 10
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin ๐ฅ:
2๐+1
๐ฅ
3
5
๐ฅ โ ๐ฅ3! + ๐ฅ5! + โฏ + โ1 ๐
2๐ + 1 !
Investigate the partial sums of the series and compare the
results to ๐ = sin ๐ฅ on a graph.
๐๐ = sin ๐
๐๐ = ๐
๐ฅ3
๐๐ = ๐ฅ โ 3!
๐๐ = ๐ฅ โ
๐ฅ3
๐ฅ5
3! + 5!
๐ฅ3
๐ฅ5
๐ฅ7
๐๐ = ๐ฅ โ 3! + 5! โ 7!
๐ฅ3
๐ฅ5
๐ฅ7
๐ฅ9
๐๐ = ๐ฅ โ 3! + 5! โ 7! + 9!
Window: โ7 โค ๐ฅ โค 7 and โ5 โค ๐ฆ โค 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin ๐ฅ:
2๐+1
๐ฅ
3
5
๐ฅ โ ๐ฅ3! + ๐ฅ5! + โฏ + โ1 ๐
2๐ + 1 !
Investigate the partial sums of the series and compare the
results to ๐ = sin ๐ฅ on a graph.
๐๐ = sin ๐
๐๐ = ๐
๐ฅ3
๐๐ = ๐ฅ โ 3!
๐๐ = ๐ฅ โ
๐ฅ3
๐ฅ5
3! + 5!
๐ฅ3
๐ฅ5
๐ฅ7
๐๐ = ๐ฅ โ 3! + 5! โ 7!
๐ฅ3
๐ฅ5
๐ฅ7
๐ฅ9
๐๐ = ๐ฅ โ 3! + 5! โ 7! + 9!
Window: โ7 โค ๐ฅ โค 7 and โ5 โค ๐ฆ โค 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin ๐ฅ:
2๐+1
๐ฅ
3
5
๐ฅ โ ๐ฅ3! + ๐ฅ5! + โฏ + โ1 ๐
2๐ + 1 !
Investigate the partial sums of the series and compare the
results to ๐ = sin ๐ฅ on a graph.
๐๐ = sin ๐
๐๐ = ๐
๐ฅ3
๐๐ = ๐ฅ โ 3!
๐๐ = ๐ฅ โ
๐ฅ3
๐ฅ5
3! + 5!
๐ฅ3
๐ฅ5
๐ฅ7
๐๐ = ๐ฅ โ 3! + 5! โ 7!
๐ฅ3
๐ฅ5
๐ฅ7
๐ฅ9
๐๐ = ๐ฅ โ 3! + 5! โ 7! + 9!
Window: โ7 โค ๐ฅ โค 7 and โ5 โค ๐ฆ โค 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin ๐ฅ:
2๐+1
๐ฅ
3
5
๐ฅ โ ๐ฅ3! + ๐ฅ5! + โฏ + โ1 ๐
2๐ + 1 !
Investigate the partial sums of the series and compare the
results to ๐ = sin ๐ฅ on a graph.
๐๐ = sin ๐
๐๐ = ๐
๐ฅ3
๐๐ = ๐ฅ โ 3!
๐๐ = ๐ฅ โ
๐ฅ3
๐ฅ5
3! + 5!
๐ฅ3
๐ฅ5
๐ฅ7
๐๐ = ๐ฅ โ 3! + 5! โ 7!
๐ฅ3
๐ฅ5
๐ฅ7
๐ฅ9
๐๐ = ๐ฅ โ 3! + 5! โ 7! + 9!
Window: โ7 โค ๐ฅ โค 7 and โ5 โค ๐ฆ โค 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin ๐ฅ:
2๐+1
๐ฅ
3
5
๐ฅ โ ๐ฅ3! + ๐ฅ5! + โฏ + โ1 ๐
2๐ + 1 !
Investigate the partial sums of the series and compare the
results to ๐ = sin ๐ฅ on a graph.
๐๐ = sin ๐
๐๐ = ๐
๐ฅ3
๐๐ = ๐ฅ โ 3!
๐๐ = ๐ฅ โ
๐ฅ3
๐ฅ5
3! + 5!
๐ฅ3
๐ฅ5
๐ฅ7
๐๐ = ๐ฅ โ 3! + 5! โ 7!
๐ฅ3
๐ฅ5
๐ฅ7
๐ฅ9
๐๐ = ๐ฅ โ 3! + 5! โ 7! + 9!
Window: โ7 โค ๐ฅ โค 7 and โ5 โค ๐ฆ โค 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin ๐ฅ:
2๐+1
๐ฅ
3
5
๐ฅ โ ๐ฅ3! + ๐ฅ5! + โฏ + โ1 ๐
2๐ + 1 !
Investigate the partial sums of the series and compare the
results to ๐ = sin ๐ฅ on a graph.
๐๐ = sin ๐
๐๐ = ๐
๐ฅ3
๐๐ = ๐ฅ โ 3!
๐๐ = ๐ฅ โ
๐ฅ3
๐ฅ5
3! + 5!
๐ฅ3
๐ฅ5
๐ฅ7
๐๐ = ๐ฅ โ 3! + 5! โ 7!
๐ฅ3
๐ฅ5
๐ฅ7
๐ฅ9
๐๐ = ๐ฅ โ 3! + 5! โ 7! + 9!
Window: โ7 โค ๐ฅ โค 7 and โ5 โค ๐ฆ โค 5
When is a Taylor Series a Good
Approximation?
Consider the Maclaurin Series for sin ๐ฅ:
2๐+1
๐ฅ
3
5
๐ฅ โ ๐ฅ3! + ๐ฅ5! + โฏ + โ1 ๐
2๐ + 1 !
On (โโ, โ), {๐๐ , ๐๐ , ๐๐ , ๐๐ , ๐๐ ,โฆ} converges to ๐๐ = sin ๐. The
polynomial series is a good approximation of ๐๐ on โโ, โ .
๐๐ = sin ๐
๐๐ = ๐
๐ฅ3
๐๐ = ๐ฅ โ 3!
๐๐ = ๐ฅ โ
๐ฅ3
๐ฅ5
3! + 5!
๐ฅ3
๐ฅ5
๐ฅ7
๐๐ = ๐ฅ โ 3! + 5! โ 7!
๐ฅ3
๐ฅ5
๐ฅ7
๐ฅ9
๐๐ = ๐ฅ โ 3! + 5! โ 7! + 9!
Window: โ7 โค ๐ฅ โค 7 and โ5 โค ๐ฆ โค 5
Proving Sine Converges
Prove the Maclaurin series
โ
๐=0
โ1
Investigate the error function:
sin ๐ฅ for all real ๐ฅ.
๐ ๐ฅ โ 0 ๐+1
๐
๐ ๐ฅ =
๐+1 !
๐ ๐+1 ๐ ๐ฅ โ 0 ๐+1
๐
๐ ๐ฅ =
๐+1 !
๐
๐+1
๐ ๐ฅ ๐+1
=
๐+1 !
1 โ ๐ฅ ๐+1
๐ฅ ๐+1
โค
โค
๐+1 !
๐+1 !
๐
๐+1
2๐+1
๐ ๐ฅ
2๐+1 !
converges to
Investigate the limit of the
last statement
๐ฅ ๐+1
=0
lim
๐โโ ๐ + 1 !
This means that ๐
๐ ๐ฅ โ 0
for all ๐ฅ.
Therefore the Maclaurin
series converges for all ๐ฅ.
When is a Taylor Series a Good
Approximation?
โ1 ๐ฅ 2
Consider the Maclaurin Series for ๐
0
0
๐ฅโ 0:
๐ฅ=0
Investigate the partial sums of the series and compare the
results to ๐ =
๐๐ =
2
โ1
๐ฅ
๐
0
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐ฅโ 0
๐ฅ=0
2
โ1
๐ฅ
๐
0
๐ฅ โ 0 on a graph.
๐ฅ=0
When is a Taylor Series a Good
Approximation?
โ1 ๐ฅ 2
Consider the Maclaurin Series for ๐
0
0
๐ฅโ 0:
๐ฅ=0
Investigate the partial sums of the series and compare the
results to ๐ =
๐๐ =
2
โ1
๐ฅ
๐
0
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐ฅโ 0
๐ฅ=0
2
โ1
๐ฅ
๐
0
๐ฅ โ 0 on a graph.
๐ฅ=0
When is a Taylor Series a Good
Approximation?
โ1 ๐ฅ 2
Consider the Maclaurin Series for ๐
0
0
๐ฅโ 0:
๐ฅ=0
Investigate the partial sums of the series and compare the
results to ๐ =
๐๐ =
2
โ1
๐ฅ
๐
0
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐ฅโ 0
๐ฅ=0
2
โ1
๐ฅ
๐
0
๐ฅ โ 0 on a graph.
๐ฅ=0
When is a Taylor Series a Good
Approximation?
โ1 ๐ฅ 2
Consider the Maclaurin Series for ๐
0
0
๐ฅโ 0:
๐ฅ=0
Investigate the partial sums of the series and compare the
results to ๐ =
๐๐ =
2
โ1
๐ฅ
๐
0
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐ฅโ 0
๐ฅ=0
2
โ1
๐ฅ
๐
0
๐ฅ โ 0 on a graph.
๐ฅ=0
When is a Taylor Series a Good
Approximation?
โ1 ๐ฅ 2
Consider the Maclaurin Series for ๐
0
0
๐ฅโ 0:
๐ฅ=0
Investigate the partial sums of the series and compare the
results to ๐ =
๐๐ =
2
โ1
๐ฅ
๐
0
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐ฅโ 0
๐ฅ=0
2
โ1
๐ฅ
๐
0
๐ฅ โ 0 on a graph.
๐ฅ=0
When is a Taylor Series a Good
Approximation?
โ1 ๐ฅ 2
Consider the Maclaurin Series for ๐
0
0
๐ฅโ 0:
๐ฅ=0
Investigate the partial sums of the series and compare the
results to ๐ =
๐๐ =
2
โ1
๐ฅ
๐
0
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐ฅโ 0
๐ฅ=0
2
โ1
๐ฅ
๐
0
๐ฅ โ 0 on a graph.
๐ฅ=0
When is a Taylor Series a Good
Approximation?
โ1 ๐ฅ 2
Consider the Maclaurin Series for ๐
0
0
On ๐ , {๐๐ , ๐๐ , ๐๐ , ๐๐ , ๐๐ ,โฆ} converges to ๐๐ =
๐ฅโ 0:
๐ฅ=0
2
โ1
๐ฅ
๐
0
The series is a good approximation of ๐๐ on ๐ .
๐๐ =
2
โ1
๐ฅ
๐
0
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐๐ = ๐
๐ฅโ 0
๐ฅ=0
๐ฅ โ 0.
๐ฅ=0
The Convergence Theorem for Power
Series
There are three possibilities for
respect to convergence:
โ
๐=0 ๐๐
๐ฅโ๐
๐
with
Radius of Convergence
1. There is a positive number ๐
such that the series
diverges for ๐ฅ โ ๐ > ๐
but converges for ๐ฅ โ ๐ <
๐
. The series may or may not converge at either of
the endpoints ๐ฅ = ๐ โ ๐
and ๐ฅ = ๐ + ๐
.
2. The series converges for every ๐ฅ (๐
= โ).
3. The series converges at ๐ฅ = ๐ and diverges
elsewhere (๐
= 0).
The set of all values of ๐ for which the series converges is the
interval of convergence.
The n-th Term Test
If lim ๐๐ โ 0, then the infinite series
๐โโ
โ
๐=1 ๐๐
diverges.
OR
When
determining if a
series converges,
always use this
test first!
If the infinite series โ
๐=1 ๐๐ converges,
then lim ๐๐ = 0.
๐โโ
The converse of this statement is NOT true.
If lim ๐๐ = 0, the infinite series
โ
๐=1 ๐๐
does not necessarily converge.
Examples
Use the n-th Term Test to investigate the convergence of
the series.
1.
โ 2๐+1
๐=1 ๐
Since the limit is not 0,
๐๐ + ๐
=๐
lim
the series must diverge.
๐โโ
๐
2.
5
โ
๐=1 10๐
๐
=๐
lim
๐
๐โโ ๐๐
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 ๐๐ = 0?
๐โโ
Yes or Maybe
?
No
The series
diverges.
Convergent Geometric Series
๐โ1
The geometric series โ
๐๐
converges
๐=1
if and only if ๐ < 1. If the series converges,
๐
its sum is
.
1โ๐
Where a is the first term and r
is the constant ratio.
Example: Determine if
๐ ๐
โ
(
๐=1 ๐)
converges.
๐
๐
This is a geometric series with ๐ = .
Since
๐
๐
< ๐, the series converges.
This explanation would warrant full credit on the AP Test. Statements like โI
know this is a __ series with __, so it diverges/convergesโ are acceptable.
The Finney Procedure for Determining
Convergence
nth-Term Test
Is lim ๐๐ = 0?
The series
diverges.
No
๐โโ
Yes or Maybe
Geometric Series Test
Is ๐๐ = ๐ + ๐๐ + ๐๐ 2 + โฏ ?
No
?
Yes
Converges to
๐
if ๐ < 1.
1โ๐
Diverges if
๐ > 1.
Example
๐! 2
โ
๐=0 2๐ !
Investigate the series
to see if it
diverges.
Notice the series is NOT geometric.
โ
๐=0
๐! 2
1 1 1
1
1
=1+ + +
+
+
+โฏ
2๐ !
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
๏ฝ
1 2
a3
๏ฝ 0.5;
1
a2
a1
๏ฝ
1 6
1 2
a2
๏ฝ 0.33;
๏ฝ
1 20
a5
๏ฝ 0.3;
1 6
a4
a3
๏ฝ
1 70
a4
๏ฝ
1 252
๏ฝ 0.28
1 70
๏ฝ 0.29;
1 20
Similar to a converging Geometric Series, the limit of the โcommonโ ratio appears to be
less than 1. According to the next theorem. This means the series converges.
The Ratio Test
Suppose the limit
๐๐+1
lim
๐โโ ๐๐
= ๐ฟ either exists
or is infinite. Then:
1. If ๐ฟ < 1, the series
โ
๐=1 ๐๐ converges.
Each successive terms are getting smaller.
โ
๐=1 ๐๐ diverges.
Each successive terms are getting larger.
2. If ๐ฟ > 1, the series
3. If ๐ฟ = 1, the test is inconclusive.
This is the โcousinโ test to determining if a Convergent Geometric Series
test. It is identical except: (1) If the โcommonโ ratio is 1, then test fails. (2)
The test does not determine the value the series converges to.
Example 1
Investigate the series
๐! 2
โ
๐=0 2๐ !
to see if it diverges.
Similar to a Geometric Series, we can investigate the ratio between terms.
๏จ ๏จ n ๏ซ 1 ๏ฉ !๏ฉ
๏จ 2 ๏จ n ๏ซ1๏ฉ ๏ฉ!
2
lim
n๏ฎ ๏ฅ
a n ๏ซ1
๏ฝ lim
an
๏ฝ lim
n๏ฎ ๏ฅ
๏จ n !๏ฉ
๏จ 2 n ๏ฉ!
2
n๏ฎ ๏ฅ
๏จ n ๏ซ1๏ฉ
2
๏จ 2 n ๏ซ 2 ๏ฉ๏จ 2 n ๏ซ 1 ๏ฉ
๏จ ๏จ n ๏ซ 1 ๏ฉ !๏ฉ ๏จ 2 n ๏ฉ !
2
๏ฝ lim
๏จ n !๏ฉ ๏จ 2 n ๏ซ 2 ๏ฉ !
2
n๏ฎ ๏ฅ
๏ฝ lim
2
n ๏ซ 2 n ๏ซ1
2
n๏ฎ ๏ฅ 4n ๏ซ6n๏ซ2
๏ฝ
1
4
Similar to a converging Geometric Series, this โcommonโ 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
โ
2๐ . Find the radius of convergence.
๐ฅ
โ
2
๐=1 32๐ 2๐
If the series
1
lim
n๏ฎ ๏ฅ
a n ๏ซ1
an
๏ฝ
3
lim
2 ๏จ n ๏ซ1๏ฉ
๏จ 2 ๏จ n ๏ซ 1๏ฉ ๏ฉ
1
n๏ฎ ๏ฅ
3
๏ฝ lim
n๏ฎ ๏ฅ
2n
๏จ 2n ๏ฉ
2n๏ซ2
๏จ x ๏ญ 2๏ฉ
๏จ 2n ๏ฉ ๏จ x ๏ญ 2 ๏ฉ
2n
2n๏ซ2
3
๏จ 2 ๏จ n ๏ซ 1๏ฉ ๏ฉ ๏จ x ๏ญ 2 ๏ฉ
3
2n
๏จ x ๏ญ 2๏ฉ
2n
2 ๏จ n ๏ซ1๏ฉ
converges, the
ratio is less
than 1.
๏จ x๏ญ2๏ฉ
2
9
๏จ x ๏ญ 2๏ฉ
2
๏ผ1
๏ผ9
x๏ญ2 ๏ผ3
The interval of
convergence is
The limit depends on n. Separate the nโs.
2
2
2 centered at 2 with
n ๏จ x ๏ญ 2๏ฉ
๏จ x ๏ญ 2๏ฉ
n
๏จ x ๏ญ 2๏ฉ
๏
๏ฝ 1๏
a radius
lim 2
๏ฝ lim
2
n๏ฎ ๏ฅ 3
3
9
๏จ n ๏ซ 1๏ฉ n ๏ฎ ๏ฅ n ๏ซ 1
convergence of 3.
Example 3
Find the radius of convergence for the Maclaurin series for
2
๐ ๐ฅ = ๐๐ฅ .
โ
2
3
๐
๐ฅ
๐ฅ
๐ฅ
We know: ๐ ๐ฅ = 1 + ๐ฅ + + + โฏ + + โฏ =
2! 3!
๐!
Thus: ๐
๐ฅ2
4
6
2๐
๐ฅ
๐ฅ
๐ฅ
= 1 + ๐ฅ2 + + + โฏ +
+โฏ=
2! 3!
๐!
Investigate
the ratio:
lim
n๏ฎ ๏ฅ
a n ๏ซ1
an
๐=0
โ
x
๏ฝ lim
n๏ฎ ๏ฅ
๐ฅ๐
๐!
๐=0
๐ฅ 2๐
๐!
2 ๏จ n ๏ซ1๏ฉ
๏จ n ๏ซ 1๏ฉ !
x
2n
๏ฝ lim
n๏ฎ ๏ฅ
x
2n๏ซ2
n!
๏จ n ๏ซ 1๏ฉ ! x
2n
๏ฝ lim
n๏ฎ ๏ฅ
n!
The series has an infinite radius.
x
2
n ๏ซ1
๏ฝ 0
Example 4
Find the radius of convergence for the series
๏จ ๏ญ 1๏ฉ
a n ๏ซ1
Investigate
lim
๏ฝ lim
the ratio: n ๏ฎ ๏ฅ a
n๏ฎ ๏ฅ
n
๏จ n ๏ซ 1๏ฉ ๏ญ1
5
x
n ๏ซ1
n ๏ซ1
๏จ ๏ญ 1๏ฉ
n ๏ญ1
5
x
โ1 ๐โ1 ๐ฅ ๐
โ
.
๐=1
๐
5
n
๏ฝ lim
n๏ฎ ๏ฅ
5
5
n
๏จ ๏ญ 1๏ฉ x
n ๏ซ1
n
๏จ ๏ญ 1๏ฉ
n ๏ซ1
n ๏ญ1
n
Since the ratio tests uses
n
n ๏ซ1
5 x
x
x
an absolute value, the
๏ฝ lim
๏ฝ
powers on -1 do not ๏ฝ lim
n ๏ซ1 n
n๏ฎ ๏ฅ 5
n๏ฎ ๏ฅ 5
x
5
affect the limit.
If the series
converges, the
ratio is less
than 1.
x
5
๏ผ1
x ๏ผ5
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
โ ๐+1
convergence for ๐=1 ๐ ๐ฅ.
lim
n๏ฎ ๏ฅ
a n ๏ซ1
an
๏จ n ๏ซ1๏ฉ ๏ซ1
๏ฝ lim
n๏ฎ ๏ฅ
n ๏ซ1
n ๏ซ1
n
x
x
๏ฝ lim
n๏ฎ ๏ฅ
n๏จn๏ซ 2๏ฉ
n ๏จ n ๏ซ1๏ฉ
๏ฝ lim
n๏ฎ ๏ฅ
n๏ซ2
n ๏ซ1
๏ฝ 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 โค ๐๐ โค ๐๐ for all ๐ โฅ ๐.
Both series must have only positive terms.
โข If
โข If
โ
๐=1 ๐๐
โ
๐=1 ๐๐
converges, then โ
๐=1 ๐๐ converges.
diverges, then โ
๐=1 ๐๐ diverges.
Example: Prove
For all ๐: 0 โค
2๐
๐ฅ 2๐
๐! 2
๐ฅ 2๐
โ
๐=0 ๐! 2
โค
๐ฅ 2๐
๐!
converges for all real ๐ฅ.
๐ฅ 2๐
โ
๐=0 ๐! is the Taylor
Series for ๐ ๐ฅ =
โ ๐ฅ
Since ๐=0
converges for all real ๐,
๐!
2
๐ฅ
๐ .
๐ฅ 2๐
โ
๐=0 ๐! 2 converges
all real ๐ by the Direct Comparison Test.
for
What about Negative Terms?
Does the Direct Comparison Test fail if there are negative terms?
Consider the
sin ๐ฅ ๐
โ
series ๐=0
๐!
โ
๐
sin 3๐
2
๐=0
๐!
if ๐ฅ = 3๐
:
2
1
1
= 1 โ 1 + 12 โ 16 + 24
โ 120
โฆ
Using partial sums, it appears the series converges:
1; 0; 0.5; 0.333; 0.375; 0.367
But we donโt 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:
โ
๐
sin 3๐
2
1
1
1
= 1 + 1 + 12 + 16 + 24
+ 120
โฆ + ๐!
๐!
๐=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
โ
๐=1 ๐๐ converges, then
โ
๐=1 ๐๐ converges.
Unlike the Comparison Test, this test does not require the terms
to be positive.
Example: Prove
Investigate
sin ๐ฅ ๐
๐!
sin ๐ฅ ๐
โ
๐=0
๐!
Notice:
sin ๐ฅ ๐
1
โค
๐!
๐!
converges for all ๐ฅ.
โ 1
๐=0 ๐!
lim
n๏ฎ ๏ฅ
a n ๏ซ1
an
๏ฝ
converges by the Ratio Test:
1
lim
n๏ฎ ๏ฅ
๏จ n ๏ซ 1๏ฉ !
1
๏ฝ lim
n๏ฎ ๏ฅ
n!
๏จ n ๏ซ 1๏ฉ !
๏ฝ lim
n๏ฎ ๏ฅ
1
n ๏ซ1
n!
sin ๐ฅ ๐
Since
converges for
๐!
sin ๐ฅ ๐
โ
converges for all
๐=0
๐!
all real ๐ฅ by the Direct Comparison Test,
real ๐ฅ by the Absolute Convergence Test.
๏ฝ
0
Definition of Absolute Convergence
If โ
๐=1 ๐๐ converges, then
absolutely.
โ
๐=1 ๐๐
converges
Example: The Alternating Harmonic Series
converges but it does not converge absolutely:
โ
(โ1)๐โ1
1
๐
= 1 โ 12 + 13 โ 14 + โฏ = ln 2
๐=1
โ
(โ1)๐โ1
๐=1
โ
1
๐
1
๐
=
๐=1
= 1 + 12 + 13 + 14 + โฏ
This is the divergent
Harmonic Series.
Note about Absolute Convergence
If โ
๐=1 ๐๐ converges, then
absolutely.
โ
๐=1 ๐๐
converges
๐๐+1
๐๐
The Ratio Test uses an absolute value:
.
Thus, every test that converges by the Ratio
Test also converges absolutely:
โ
๐โ1
4(โ0.5)
๐=1
โ
๐โ1
๐=1 4(0.5)
converges absolutely because
converges by by the ratio test
The Ratio Test is incredibly strong