Interval of Convergence: Integral
Test, p-Series, Alternating Series
Test, Alternating Series Error
Using an Integral to Determine Convergence
Determine if the Harmonic Series converges:
∞
1
𝑛
= 1 + 12 + 13 + 14 + ⋯
𝑛=1
Compare the Series to the
𝟏
graph of 𝒚 = .
𝒙
Find the Left Hand Riemann
∞𝟏
Sum to approximate 𝟏 𝒅𝒙.
𝒙
1     ...
1
2
…
…
1
3
1
4
1
5
The Left Hand Riemann Sum
is equal to the Sum of the
Harmonic Series.
Using an Integral to Determine Convergence
Determine if the Harmonic Series converges:
∞
1
𝑛
= 1 + 12 + 13 + 14 + ⋯
𝑛=1
Compare the Series to the
𝟏
graph of 𝒚 = .
𝒙
So…
∞ 1
𝑛=1 𝑛 = 𝑇ℎ𝑒 𝐿𝐻𝑅𝑆 𝑜𝑓
∞1
𝑑𝑥
1 𝑥
𝟏
Since 𝒚 = is decreasing, the Left
𝒙
Hand Riemann Sum is an over
estimate.
Thus:
…
∞
𝑳𝒆𝒇𝒕 𝑯𝒂𝒏𝒅 𝑹𝒊𝒆𝒎𝒂𝒏𝒏 𝑺𝒖𝒎 >
𝟏
𝟏
𝒅𝒙
𝒙
Using an Integral to Determine Convergence
Determine if the Harmonic Series converges:
∞
1
𝑛
= 1 + 12 + 13 + 14 + ⋯
∞𝟏
∞ 1
𝑛=1
So… 𝑛=1 𝑛 > 𝟏 𝒅𝒙
𝒙
Compare the Series to the
We can find the value of the
𝟏
graph of 𝒚 = .
improper integral:
𝒙
lim 
b
1
x
b  1
…
dx  lim  ln x  1
b
b 
 lim  ln b  ln1
b 

∞𝟏
1
Since 𝟏 𝒅𝒙 diverges and ∞
𝑛=1 𝑛
𝒙
∞𝟏
𝒅𝒙, the Harmonic Series
𝟏 𝒙
Diverges.
Using an Integral to Determine Convergence
Determine if the series below converges:
∞
1
𝑛2
1
= 1 + 14 + 19 + 16
+⋯
𝑛=1
Compare the Series to the
𝟏
graph of 𝒚 = 𝟐 .
𝒙
Find the Right Hand Riemann
∞𝟏
Sum to approximate 𝟏 𝒅𝒙.
𝒙
1   
1
4
…
…
1
9
1
16
1
25
 ...
The right Hand Riemann Sum
is equal to the Sum of the
Harmonic Series.
Using an Integral to Determine Convergence
Determine if the series below converges:
∞
1
𝑛2
1
= 1 + 14 + 19 + 16
+⋯
𝑛=1
Compare the Series to the
𝟏
graph of 𝒚 = 𝟐 .
𝒙
1
∞
𝑛=1 𝑛2
So…
= 𝑇ℎ𝑒 𝑅𝐻𝑅𝑆 𝑜𝑓
∞ 1
𝑑𝑥
1 𝑥2
𝟏
Since𝒚 = 𝟐 is decreasing, the Right
𝒙
Hand Riemann Sum is an under
estimate.
Thus:
…
…
∞
𝑹𝒊𝒈𝒉𝒕 𝑯𝒂𝒏𝒅 𝑹𝒊𝒆𝒎𝒂𝒏𝒏 𝑺𝒖𝒎 <
𝟏
𝟏
𝒅𝒙
𝒙
Using an Integral to Determine Convergence
Determine if the series below converges:
∞
1
𝑛2
1
= 1 + 14 + 19 + 16
+⋯
∞ 𝟏
1
∞
𝑛=1
So… 𝑛=1 𝑛2 < 𝟏 𝟐 𝒅𝒙
𝒙
Compare the Series to the
We can find the value of the
𝟏
graph of 𝒚 = 𝟐 .
improper integral:
𝒙
Extension: Since the
area is an over estimate,
we have a bound for the
value of our series
∞
1
𝑛2
𝑛=1
…
…
<1
lim 
b
1
2
b  1 x
dx  lim  

1
b
 lim   b1  11 
b
1
1
x
b
∞ 𝟏
Since 𝟏 𝟐 𝒅𝒙 converges and
𝒙
∞ 𝟏
1
∞
<
𝒅𝒙, the series
𝑛=1 𝑛2
𝟏 𝒙𝟐
converges.
The Integral Test
If 𝑓 𝑥 is continuous, positive, and
decreasing for all 𝑥 ≥ 𝑀 > 0, then ∞
𝑛=1 𝑓 𝑛
∞
converges if and only if 𝑀 𝑓 𝑥 𝑑𝑥
converges.
You can begin the summation at 𝒏 = 𝟏 instead of 𝑴 because
you can remove any finite number of beginning terms of a
series without affecting its convergence or divergence.
Extension: If
∞
𝑛=1 𝑓
𝑛 converges and S =
∞
𝑓 𝑥 𝑑𝑥.
𝑀
∞
𝑛=𝑀 𝑓
𝑛 , then S <
Example
𝑛
∞
𝑛=1 (𝑛2 +1)2
Does
converge?
The function is positive, continuous, and decreasing.
Therefore the Integral Test applies:


1
1
2
x
 x 1
2

2 dx  lim 
b  1
lim 
b 

1
4
b

b
1
x 2 1 1
x
 x 1
2

2 dx 
 lim 
1
2
b 
1
b 2 1
1
2

lim 
b
b  1
1
12 1
2x
 x 1
2
2
dx
  0  
Since the integral converges, the series
also converges.
1
2
1
2
White Board Challenge
Consider the series below:
∞
𝑛=1
1
𝑛𝑝
What values of 𝑝 does the series converge? Diverge?
𝑝 < 1: Diverges by the nth-Term test. For example:
1
∞
∞
10 diverges since lim 𝑛10 ≠ 0.
=
𝑛
𝑛=1 −10
𝑛=1
𝑛
𝑛→∞
𝑝 = 1: Diverges since it is the Harmonic Series:
1
1
1
1
∞
𝑛=1 1 = 1 + 2 + 3 + 4 + ⋯.
𝑛
𝑝 < 1: Converges by the Integral test. For example:
The p-Series Test
A p-Series is any series of the form
1
∞
𝑛=1 𝑛𝑝 .
1
∞
𝑛=1 𝑛𝑝
converges if and only if 𝑝 > 1.
This is a direct result of the Integral Test.
1
∞
𝑛=1 𝑛0.99
Consider this statement:
diverges but
converges.
1
∞
𝑛=1 𝑛1.01
The Inability to Find the Sum of a Series
Like every test for convergence, except for the Geometric
Series Test, the p-Series test says the series below
converges but does say what the infinite sum equals.


1
1.01
n
n 1
But, since the p-Series test relies on the integral test, we
can determine an upper bound for the series above:


1
1.01
n 1 n
 1


1
x 1.01
dx  lim
b

b
1
1.01
1 x
dx  lim 100x
b
 lim 100b0.01  100 10.01  100
b
0.01



b
1
Thus:
1
n 1 n
1.01
100
Example 1
Prove
∞ sin2 𝑛
𝑛=1 𝑛2
converges.
For all 𝑛:
1
∞
𝑛=1 𝑛2
sin2 𝑛
𝑛2
≤
1
𝑛2
is a convergent p-Series since 𝑝 = 2
(greater than 1).
Since
1
∞
𝑛=1 𝑛2
converges, by the Direct
Comparison Test
∞ sin2 𝑛
𝑛=1 𝑛2
converges.
Example 2
Prove
∞ ln 𝑛
𝑛=1 𝑛
diverges.
For all 𝑛 > 2: ln𝑛𝑛 > 𝑛1
Notice: All we need is for the terms to eventually
get larger than the terms of the known series.
∞ 1
𝑛=1 𝑛
is a divergent p-Series since 𝑝 = 1 (not greater
than 1).
Since
∞ 1
𝑛=1 𝑛
diverges, by the Comparison Test
∞ ln 𝑛
𝑛=1 𝑛
diverges.
The Finney Procedure for Determining
Convergence
No
nth-Term Test
Is lim 𝑎𝑛 = 0?
The series diverges.
𝑛→∞
Yes or Maybe
Is
Geometric Series Test
𝑎𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + ⋯ ?
Yes
𝑎
Converges to 1−𝑟
if 𝑟 < 1.
Diverges if 𝑟 > 1.
No
p-Series Test
Does the series have the form
1
∞
𝑛=1 𝑝 ?
Yes
𝑛
Converges if 𝑝 >
1.
Diverges if 𝑝 ≤ 1.
No or maybe
Does 𝑎𝑛 converge?
Apply the Comparison
test, Integral Test, or
Ratio Test
…
Yes
The original
Series
Converges.
An Alternating Series that Converges
Justify the Alternating Harmonic Series converges:
∞
1
(−1)𝑛−1 𝑛
= 1 − 12 + 13 − 14 + ⋯
𝑛=1
Investigate and plot the sum:
1
1
1
1
1
1
7
5
6
2
3
4
1       ...
Each Successive term
in the sequence of partial
sums is between the two
previous terms in
this sequence .
0
The sum is bounded by 1
0.5 and 1.
2
S ?
The sum must be between
any two successive terms.
1
An Alternating Series that Diverges
Justify the Alternating Harmonic Series converges:
∞
(−1)𝑛−1 𝑛
=1−2+3−4+⋯
𝑛=1
Investigate and plot the sum:
Graph
the
partial
sums:
1 2  3  4  5  6 
The Series
Diverges
Each Successive
term in the sequence of
partial sums is outside the
two previous terms
in this sequence .
In order to converge; the terms must be alternating, if you take the
absolute value of each term the terms must be decreasing, and the nth
term must approach 0.
The Alternating Series Test
If the following is true:
1. 𝑎𝑛 > 0
2. 𝑎𝑛 is decreasing
3. lim 𝑎𝑛 = 0
𝑛→∞
Then
∞
𝑛=1
−1
𝑛−1
𝑎𝑛 and
converge.
∞
𝑛=1
−1
𝑛
𝑎𝑛
Example 1
𝑛+1 𝑛2
∞
𝑛=1
Test the series
−1
for convergence or
𝑛3 +1
divergence.
The Series COULD be alternating because it has the −1
Now check the conditions:
1: Since 𝑛 ≥ 1, 𝑎𝑛 =
2. We know
𝑑
𝑑𝑛
𝑎𝑛 =
𝑛2
𝑛3 +1
𝑛→∞
>0
𝑛(2−𝑛2 )
𝑛3 +1 2
3. We know lim 𝑎𝑛 =
𝑛+1 .
𝑛2
𝑛3 +1
< 0 when 𝑛 > 0 so 𝑎𝑛 is decreasing.
=0
Since the Series satisfies all of the conditions of the Alternating
Series Test, it must converge.
Example 2
Test the series
−1 𝑛 3𝑛
∞
𝑛=1 4𝑛−1
for convergence or divergence.
The Series COULD be alternating because it has the −1 𝑛 . Now
check the conditions:
1: Since 𝑛 ≥ 1, 𝑎𝑛 =
2. We know
𝑑
𝑑𝑛
𝑎𝑛 =
3𝑛
4𝑛−1
−3
4𝑛−1 2
3. BUT know lim 𝑎𝑛 =
𝑛→∞
Since
−1 𝑛 3𝑛
lim
𝑛→∞ 4𝑛−1
>0
< 0 for all 𝑛 so 𝑎𝑛 is decreasing.
3𝑛
4𝑛−1
= 34 ≠ 0
This means the Series
diverges by the nth-Term
Test. We should have
checked this first!
does not equal 0, the series must DIVERGE.
Error for an Alternating Series
If the first three terms are used to estimate the convergent Alternating
Harmonic series below, estimate the error of the approximation:
∞
1
(−1)𝑛−1 𝑛
𝑆=
𝑛=1
Investigate and plot the sum
of the first three terms:
= 1 − 12 + 13 − 14 + ⋯
1 12 
…and the first
three terms are used to
approximate the infinite
Alternating Series...
1
3
1
4
 S  1  12  13 
Actual Error
0
S
If the series converges… 1
2 …the error of the approximation
1
can not be larger than the first
omitted term.
The Alternating Series Estimation Theorem
𝑛−1
Let 𝑆 = ∞
−1
𝑎𝑛 , where 𝑎𝑛 is a
𝑛=1
positive decreasing sequence that
converges to 0. Then
𝑆 − 𝑆𝑛 < 𝑎𝑛+1
In other words, the error committed when we
approximate a convergent Alternating series
with the first n terms is less than the size of
the first omitted term.
Notice from the diagram on the last slide, that this Theorem
tells the magnitude AND the direction of the error!
Example
Consider S =
−1 𝑛−1
∞
.
𝑛=1
𝑛
a) Prove the series converges.
You can use the Alternating Series test, but if we investigate
the series:
∞
𝑛=1
−1 𝑛−1
𝑛
=1−
1
1
1
+3−4
2
+⋯
We should notice that it is the Alternating Harmonic Series.
Thus, we already know that it converges.
Example Continued
Consider S =
−1 𝑛−1
∞
.
𝑛=1
𝑛
b) Show that 𝑆 − 𝑆6 < 0.2.
This is the definition of Error.
Since the series is a convergent Alternating Series,
the Error Bound for the Alternating Series must be less then
the first omitted term:
𝑆 − 𝑆6 < 𝑎7 =
1
≈ 0.14 < 0.2
7
Thus 𝑆 − 𝑆6 < 0.2
Example Continued
Consider S =
−1 𝑛−1
∞
.
𝑛=1
𝑛
c) Find 𝑛 such that 𝑆𝑛 approximates 𝑆 with an error less 10−3 .
Use the Alternating Series Estimation Theorem:
𝑆 − 𝑆𝑛 < 𝑎𝑛+1
1
𝐸𝑟𝑟𝑜𝑟 <
𝑛+1
If we want 𝐸𝑟𝑟𝑜𝑟 < 10−3 , then:
1
≤ 10−3
𝑛+1
→ 103 ≤ 𝑛 + 1
→ 𝑛 ≥ 999
The Finney Procedure for Determining
Convergence
No
nth-Term Test
Is lim 𝑎𝑛 = 0?
The series diverges.
𝑛→∞
Yes or Maybe
Is
Yes
Geometric Series Test
𝑎𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + ⋯ ?
𝑎
Converges to 1−𝑟
if 𝑟 < 1.
Diverges if 𝑟 > 1.
No
p-Series Test
Does the series have the form
1
∞
𝑛=1 𝑛𝑝 ?
Yes
Converges if 𝑝 > 1.
Diverges if 𝑝 ≤ 1.
Yes
The original Series Converges.
No or maybe
Does 𝑎𝑛 converge?
Apply the Comparison test, Integral Test, or Ratio Test
No or maybe
Is
𝑎𝑛 = 𝑢1 − 𝑢2 + 𝑢3 − ⋯?
(an alternating series)
No
See what you can do with
partial sums and/or try your
calculator.
Yes
Is there an integer
𝑁 such that 𝑢𝑁 ≥
𝑢𝑁+1 ≥ ⋯?
Yes
No
Series converges if 𝑢𝑛 → 0
Series diverges if 𝑢𝑛 → 0
Power Series and Convergence
How do we analytically determine ALL of the
values of x that make a power series
converge?
∞
𝑐𝑛 ∙ 𝑥 − 𝑎
𝑥=0
𝑛
How to Test a Power Series for
Convergence
1. Use the Ratio Test to find the values of 𝑥 for
which the series converges absolutely. Ordinarily,
this is an open interval: 𝑎 − 𝑅 < 𝑥 < 𝑎 + 𝑅. In
some instances, the series converges for all
values of 𝑥. In rare cases, the series
convergences only at 𝑥 = 𝑎.
2. If the interval of absolute convergence is finite,
test for convergence or divergence at each
endpoint. (The Ratio Test fails at these points.)
Use a comparison test, the Integral Test, or the
Alternating Test.
3. If the interval of absolute convergence is
𝑎 − 𝑅 < 𝑥 < 𝑎 + 𝑅, conclude that the series
diverges for 𝑥 − 𝑎 > 𝑅, because for those
values of 𝑥 the nth term does not approach zero.
Example 1
Find the interval of convergence for
1
∞
𝑛=1 32𝑛 2𝑛
We have already investigated this series:
1
2 n 1
 x  2
2 n 1
3
2  n  1 

lim
n 
1
2n
 x  2
2n
3  2n 
an 1
lim

n  a
n
𝑥−2
3
1
2
 x  2  9
x2 3
1  x  5
9
2n2
The limit depends on n. Separate the n’s.
x  2
n  x  2
n  x  2


 1
lim 2
 lim
2
n  n  1
n  3  n  1
3
9
2
2
.
If the series
converges, the ratio
is less than 1.
 x 2 2
2n  x  2 

 lim 2 n  2
2n
n 
3
 2  n  1   x  2 
2n
2𝑛
2
The Ratio test
guarantees the
series converges
on the open
interval −1,5 .
Example 1 (Continued)
1
∞
𝑛=1 32𝑛 2𝑛
Find the interval of convergence for
𝑥−2
2𝑛
.
The Ratio test guarantees the series converges on the open
interval −1,5 . But we also need to check the endpoints:
The series for
𝑥 = −1:
The series for
𝑥 = 5:
1
∞
𝑛=1 32𝑛 2𝑛
1
= 12 ∞
𝑛=1 𝑛
1
∞
𝑛=1 32𝑛 2𝑛
1
= 12 ∞
𝑛=1 𝑛
−1 − 2
2𝑛
2𝑛
=
−3
∞
𝑛=1 32𝑛 2𝑛
=
1
∞
𝑛=1 2𝑛
This is a divergent p-Series since
𝑝 = 1 (not greater than 1).
5−2
2𝑛
=
32𝑛
∞
𝑛=1 32𝑛 2𝑛
=
1
∞
𝑛=1 2𝑛
This is a divergent p-Series since
𝑝 = 1 (not greater than 1).
Since the endpoints do not produce convergent series, the series
converges on the interval −1,5 .
Example 2
2𝑛
Find the interval of convergence for the series
∞ 𝑥
𝑛=0 𝑛!
.
We have already investigated this series:
Investigate the
ratio:
an 1
lim
n  a
n
x 
n  1 !

x 2n2 n!
x2
 lim
 lim
 lim
 0
2
n
2
n
n   n  1 ! x
n 
n  n  1
x
n!
2 n 1
Therefore every value of 𝑥 produces a convergent series, thus
series converges on the interval −∞, ∞ .
Note: Convergence of Taylor Series
You should be careful when applying Taylor series to
calculating function values. Without more information, there
is no guarantee that a Taylor series converges. Moreover,
there is no guarantee, even if it does converge, that it
converges to the function that generated the series.
Consider:
The the Maclaurin Series for
𝟐
−𝟏
𝒙
𝒆
𝒙 ≠ 𝟎 is 𝑻 𝒙 = 𝟎
𝒏
𝟎
𝒙=𝟎
This Taylor Series converges for every
value of x BUT it does not converge to
the function except at 𝑥 = 0.
Example
A function 𝑓 has derivatives of all orders for all
real numbers. It is known that 𝑓 1 = 5, 𝑓′ 1 =
− 3, 𝑓′′ 1 = 2, and 𝑓′′′ 1 = −1. Which of the
following statements are necessarily true?
I.
False. The only information given concerns the value of 𝑓 and its derivatives
𝑓 2at can
exactly determined
from
given
𝑥 = 1.be
No information
can be determined
aboutthe
the value
of 𝑓 2 .
information.
True. The
We have
all we need to write
the third-degree
Taylor
for 𝑓
at 𝑥 = 1.
II.
third-degree
polynomial
for 𝑓
at 𝑥polynomial
= 1 can
be
determined from the given information.
necessarily true, since we don’t know anything about the remainder term or
III.NotThe
Taylor seriesthefor
𝑓 at 𝑥 = 1 converges to 𝑓 𝑥
interval of convergence.
for all real numbers 𝑥.
When to apply Convergence Tests
Sample of questions about convergence are…
• Which of the following series converge?
• Which of the following series diverge?
• For what values of k does a series whose nth term
is a function of k converge?
• Is it possible to evaluate f (a) with arbitrary
accuracy using its Taylor Series expansion at x =
b?
• What is the radius or interval of convergence for a
particular series?