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MAT 127 2.0 Calculus II
Dr. G.H.J. Lanel
Lecture 2
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 1 / 22
Outline
Outline
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 2 / 22
Basic Series that Converge
Outline
1 Basic Series that Converge
2 Basic Series that Diverge
3 Harmonic series
4 Necessary Condition for Convergent
5 Positive Terms Series
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 3 / 22
Basic Series that Converge
Geometric series:∞∑
n=1arn−1, if |r | < 1.
p-series:∞∑
n=1
1np , if p > 1.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 4 / 22
Basic Series that Converge
Geometric series:∞∑
n=1arn−1, if |r | < 1.
p-series:∞∑
n=1
1np , if p > 1.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 4 / 22
Basic Series that Diverge
Outline
1 Basic Series that Converge
2 Basic Series that Diverge
3 Harmonic series
4 Necessary Condition for Convergent
5 Positive Terms Series
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 5 / 22
Basic Series that Diverge
Geometric series:∞∑
n=1arn−1, if |r | ≥ 1.
p-series:∞∑
n=1
1np , if p ≤ 1.
Any series:∞∑
n=1an, for which limn→∞ an 6= 0.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 6 / 22
Basic Series that Diverge
Geometric series:∞∑
n=1arn−1, if |r | ≥ 1.
p-series:∞∑
n=1
1np , if p ≤ 1.
Any series:∞∑
n=1an, for which limn→∞ an 6= 0.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 6 / 22
Basic Series that Diverge
Geometric series:∞∑
n=1arn−1, if |r | ≥ 1.
p-series:∞∑
n=1
1np , if p ≤ 1.
Any series:∞∑
n=1an, for which limn→∞ an 6= 0.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 6 / 22
Harmonic series
Outline
1 Basic Series that Converge
2 Basic Series that Diverge
3 Harmonic series
4 Necessary Condition for Convergent
5 Positive Terms Series
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 7 / 22
Harmonic series
Definition
The series 1, 12 ,
13 ,
14 , · · · is known as Harmonic series (the p-series
when p = 1).
Claim:
Harmonic series∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 8 / 22
Harmonic series
Definition
The series 1, 12 ,
13 ,
14 , · · · is known as Harmonic series (the p-series
when p = 1).
Claim:
Harmonic series∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 8 / 22
Harmonic series
Definition
The series 1, 12 ,
13 ,
14 , · · · is known as Harmonic series (the p-series
when p = 1).
Claim:
Harmonic series∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 8 / 22
Harmonic series
Proof
Sn = 1 + 12 + 1
3 + 14 + · · ·+ 1
n
Then S1 = 1
S21=S2=1 + 12
S22=S4=1 + 12 +
(13 + 1
4
)> 1 + 1
2 +(1
4 + 14
)= 1 + 2
2
S23=S8=1 + 12 +
(13 + 1
4
)+(1
5 + 16 + 1
7 + 18
)> 1 + 1
1 +(1
4 + 14
)+(1
8 + 18 + 1
8 + 18
)= 1 + 3
2
Hence S2n > 1 + n2 for all n ∈ N.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 9 / 22
Harmonic series
Proof
Sn = 1 + 12 + 1
3 + 14 + · · ·+ 1
n
Then S1 = 1
S21=S2=1 + 12
S22=S4=1 + 12 +
(13 + 1
4
)> 1 + 1
2 +(1
4 + 14
)= 1 + 2
2
S23=S8=1 + 12 +
(13 + 1
4
)+(1
5 + 16 + 1
7 + 18
)> 1 + 1
1 +(1
4 + 14
)+(1
8 + 18 + 1
8 + 18
)= 1 + 3
2
Hence S2n > 1 + n2 for all n ∈ N.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 9 / 22
Harmonic series
Proof
Sn = 1 + 12 + 1
3 + 14 + · · ·+ 1
n
Then S1 = 1
S21=S2=1 + 12
S22=S4=1 + 12 +
(13 + 1
4
)> 1 + 1
2 +(1
4 + 14
)= 1 + 2
2
S23=S8=1 + 12 +
(13 + 1
4
)+(1
5 + 16 + 1
7 + 18
)> 1 + 1
1 +(1
4 + 14
)+(1
8 + 18 + 1
8 + 18
)= 1 + 3
2
Hence S2n > 1 + n2 for all n ∈ N.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 9 / 22
Harmonic series
Proof
Sn = 1 + 12 + 1
3 + 14 + · · ·+ 1
n
Then S1 = 1
S21=S2=1 + 12
S22=S4=1 + 12 +
(13 + 1
4
)> 1 + 1
2 +(1
4 + 14
)= 1 + 2
2
S23=S8=1 + 12 +
(13 + 1
4
)+(1
5 + 16 + 1
7 + 18
)> 1 + 1
1 +(1
4 + 14
)+(1
8 + 18 + 1
8 + 18
)= 1 + 3
2
Hence S2n > 1 + n2 for all n ∈ N.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 9 / 22
Harmonic series
Proof
Sn = 1 + 12 + 1
3 + 14 + · · ·+ 1
n
Then S1 = 1
S21=S2=1 + 12
S22=S4=1 + 12 +
(13 + 1
4
)> 1 + 1
2 +(1
4 + 14
)= 1 + 2
2
S23=S8=1 + 12 +
(13 + 1
4
)+(1
5 + 16 + 1
7 + 18
)> 1 + 1
1 +(1
4 + 14
)+(1
8 + 18 + 1
8 + 18
)= 1 + 3
2
Hence S2n > 1 + n2 for all n ∈ N.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 9 / 22
Harmonic series
Proof
Sn = 1 + 12 + 1
3 + 14 + · · ·+ 1
n
Then S1 = 1
S21=S2=1 + 12
S22=S4=1 + 12 +
(13 + 1
4
)> 1 + 1
2 +(1
4 + 14
)= 1 + 2
2
S23=S8=1 + 12 +
(13 + 1
4
)+(1
5 + 16 + 1
7 + 18
)> 1 + 1
1 +(1
4 + 14
)+(1
8 + 18 + 1
8 + 18
)= 1 + 3
2
Hence S2n > 1 + n2 for all n ∈ N.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 9 / 22
Harmonic series
Proof
Sn = 1 + 12 + 1
3 + 14 + · · ·+ 1
n
Then S1 = 1
S21=S2=1 + 12
S22=S4=1 + 12 +
(13 + 1
4
)> 1 + 1
2 +(1
4 + 14
)= 1 + 2
2
S23=S8=1 + 12 +
(13 + 1
4
)+(1
5 + 16 + 1
7 + 18
)> 1 + 1
1 +(1
4 + 14
)+(1
8 + 18 + 1
8 + 18
)= 1 + 3
2
Hence S2n > 1 + n2 for all n ∈ N.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 9 / 22
Harmonic series
Proof
Sn = 1 + 12 + 1
3 + 14 + · · ·+ 1
n
Then S1 = 1
S21=S2=1 + 12
S22=S4=1 + 12 +
(13 + 1
4
)> 1 + 1
2 +(1
4 + 14
)= 1 + 2
2
S23=S8=1 + 12 +
(13 + 1
4
)+(1
5 + 16 + 1
7 + 18
)> 1 + 1
1 +(1
4 + 14
)+(1
8 + 18 + 1
8 + 18
)= 1 + 3
2
Hence S2n > 1 + n2 for all n ∈ N.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 9 / 22
Harmonic series
Proof
Sn = 1 + 12 + 1
3 + 14 + · · ·+ 1
n
Then S1 = 1
S21=S2=1 + 12
S22=S4=1 + 12 +
(13 + 1
4
)> 1 + 1
2 +(1
4 + 14
)= 1 + 2
2
S23=S8=1 + 12 +
(13 + 1
4
)+(1
5 + 16 + 1
7 + 18
)> 1 + 1
1 +(1
4 + 14
)+(1
8 + 18 + 1
8 + 18
)= 1 + 3
2
Hence S2n > 1 + n2 for all n ∈ N.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence∞∑
n=1
1n is divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 9 / 22
Necessary Condition for Convergent
Outline
1 Basic Series that Converge
2 Basic Series that Diverge
3 Harmonic series
4 Necessary Condition for Convergent
5 Positive Terms Series
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 10 / 22
Necessary Condition for Convergent
Theorem
If∞∑
n=1an is convergent, then limn→∞ an = 0.
Proof
Suppose∞∑
n=1an is convergent.
Let Sn be the nth partial sum of the series∞∑
n=1an.
Then, an = Sn − Sn−1, for all n ∈ N.
Since the series∞∑
n=1an is convergent, limn→∞ Sn = L, for some
L ∈ R.
Hence limn→∞ an = limn→∞ (Sn − Sn−1)
= limn→∞ Sn − limn→∞ Sn−1 = L− L = 0.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 11 / 22
Necessary Condition for Convergent
Theorem
If∞∑
n=1an is convergent, then limn→∞ an = 0.
Proof
Suppose∞∑
n=1an is convergent.
Let Sn be the nth partial sum of the series∞∑
n=1an.
Then, an = Sn − Sn−1, for all n ∈ N.
Since the series∞∑
n=1an is convergent, limn→∞ Sn = L, for some
L ∈ R.
Hence limn→∞ an = limn→∞ (Sn − Sn−1)
= limn→∞ Sn − limn→∞ Sn−1 = L− L = 0.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 11 / 22
Necessary Condition for Convergent
Theorem
If∞∑
n=1an is convergent, then limn→∞ an = 0.
Proof
Suppose∞∑
n=1an is convergent.
Let Sn be the nth partial sum of the series∞∑
n=1an.
Then, an = Sn − Sn−1, for all n ∈ N.
Since the series∞∑
n=1an is convergent, limn→∞ Sn = L, for some
L ∈ R.
Hence limn→∞ an = limn→∞ (Sn − Sn−1)
= limn→∞ Sn − limn→∞ Sn−1 = L− L = 0.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 11 / 22
Necessary Condition for Convergent
Theorem
If∞∑
n=1an is convergent, then limn→∞ an = 0.
Proof
Suppose∞∑
n=1an is convergent.
Let Sn be the nth partial sum of the series∞∑
n=1an.
Then, an = Sn − Sn−1, for all n ∈ N.
Since the series∞∑
n=1an is convergent, limn→∞ Sn = L, for some
L ∈ R.
Hence limn→∞ an = limn→∞ (Sn − Sn−1)
= limn→∞ Sn − limn→∞ Sn−1 = L− L = 0.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 11 / 22
Necessary Condition for Convergent
Theorem
If∞∑
n=1an is convergent, then limn→∞ an = 0.
Proof
Suppose∞∑
n=1an is convergent.
Let Sn be the nth partial sum of the series∞∑
n=1an.
Then, an = Sn − Sn−1, for all n ∈ N.
Since the series∞∑
n=1an is convergent, limn→∞ Sn = L, for some
L ∈ R.
Hence limn→∞ an = limn→∞ (Sn − Sn−1)
= limn→∞ Sn − limn→∞ Sn−1 = L− L = 0.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 11 / 22
Necessary Condition for Convergent
Note:
If limn→∞ an = 0 does not exists or limn→∞ an 6= o, then∞∑
n=1an is
divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 12 / 22
Necessary Condition for Convergent
Note:
If limn→∞ an = 0 does not exists or limn→∞ an 6= o, then∞∑
n=1an is
divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 12 / 22
Necessary Condition for Convergent
Note:
If limn→∞ an = 0 does not exists or limn→∞ an 6= o, then∞∑
n=1an is
divergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 12 / 22
Positive Terms Series
Outline
1 Basic Series that Converge
2 Basic Series that Diverge
3 Harmonic series
4 Necessary Condition for Convergent
5 Positive Terms Series
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 13 / 22
Positive Terms Series
Definition
The series∞∑
n=1an is said to be a positive terms series, if an > 0 for all
n ∈ N.
TheoremA positive terms series convergent if and only if the sequence of
partial sums is bounded above.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 14 / 22
Positive Terms Series
Definition
The series∞∑
n=1an is said to be a positive terms series, if an > 0 for all
n ∈ N.
TheoremA positive terms series convergent if and only if the sequence of
partial sums is bounded above.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 14 / 22
Positive Terms Series
Proof: (⇐)
Let∞∑
n=1an be a positive terms series and {Sn}∞n=1 be the sequence of
partial sums.
Suppose {Sn}∞n=1 is bounded above · · · (I)
Since an > 0 for all n ∈ N, {Sn}∞n=1 is a monotonically increasingsequence · · · (II)
By (I) and (II), {Sn}∞n=1 is a convergent sequence.
Hence∞∑
n=1an is convergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 15 / 22
Positive Terms Series
Proof: (⇐)
Let∞∑
n=1an be a positive terms series and {Sn}∞n=1 be the sequence of
partial sums.
Suppose {Sn}∞n=1 is bounded above · · · (I)
Since an > 0 for all n ∈ N, {Sn}∞n=1 is a monotonically increasingsequence · · · (II)
By (I) and (II), {Sn}∞n=1 is a convergent sequence.
Hence∞∑
n=1an is convergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 15 / 22
Positive Terms Series
Proof: (⇐)
Let∞∑
n=1an be a positive terms series and {Sn}∞n=1 be the sequence of
partial sums.
Suppose {Sn}∞n=1 is bounded above · · · (I)
Since an > 0 for all n ∈ N, {Sn}∞n=1 is a monotonically increasingsequence · · · (II)
By (I) and (II), {Sn}∞n=1 is a convergent sequence.
Hence∞∑
n=1an is convergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 15 / 22
Positive Terms Series
Proof: (⇐)
Let∞∑
n=1an be a positive terms series and {Sn}∞n=1 be the sequence of
partial sums.
Suppose {Sn}∞n=1 is bounded above · · · (I)
Since an > 0 for all n ∈ N, {Sn}∞n=1 is a monotonically increasingsequence · · · (II)
By (I) and (II), {Sn}∞n=1 is a convergent sequence.
Hence∞∑
n=1an is convergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 15 / 22
Positive Terms Series
Proof: (⇐)
Let∞∑
n=1an be a positive terms series and {Sn}∞n=1 be the sequence of
partial sums.
Suppose {Sn}∞n=1 is bounded above · · · (I)
Since an > 0 for all n ∈ N, {Sn}∞n=1 is a monotonically increasingsequence · · · (II)
By (I) and (II), {Sn}∞n=1 is a convergent sequence.
Hence∞∑
n=1an is convergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 15 / 22
Positive Terms Series
Proof: (⇐)
Let∞∑
n=1an be a positive terms series and {Sn}∞n=1 be the sequence of
partial sums.
Suppose {Sn}∞n=1 is bounded above · · · (I)
Since an > 0 for all n ∈ N, {Sn}∞n=1 is a monotonically increasingsequence · · · (II)
By (I) and (II), {Sn}∞n=1 is a convergent sequence.
Hence∞∑
n=1an is convergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 15 / 22
Positive Terms Series
Example:∞∑
n=1
1n!
Let an = 1n! , for all n ∈ N and Sn be the nth partial sum of the series.
a1 = 11! = 1
a2 = 12! =
12
a3 = 13! =
11·2·3 < 1
22
an = 1n! =
11·2·3·····n < 1
2n−1
Hence Sn < 1 + 12 + 1
22 + · · ·+ 12n−1 =
1−( 12)
n
1− 12
= 2(1− 1
2n
)< 2
Hence {Sn}∞n=1 is bounded above.
Since an = 1n! > 0, for all n ∈ N and {Sn}∞n=1 is convergent,
∞∑n=1
1n! is
cgt.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 16 / 22
Positive Terms Series
Example:∞∑
n=1
1n!
Let an = 1n! , for all n ∈ N and Sn be the nth partial sum of the series.
a1 = 11! = 1
a2 = 12! =
12
a3 = 13! =
11·2·3 < 1
22
an = 1n! =
11·2·3·····n < 1
2n−1
Hence Sn < 1 + 12 + 1
22 + · · ·+ 12n−1 =
1−( 12)
n
1− 12
= 2(1− 1
2n
)< 2
Hence {Sn}∞n=1 is bounded above.
Since an = 1n! > 0, for all n ∈ N and {Sn}∞n=1 is convergent,
∞∑n=1
1n! is
cgt.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 16 / 22
Positive Terms Series
Example:∞∑
n=1
1n!
Let an = 1n! , for all n ∈ N and Sn be the nth partial sum of the series.
a1 = 11! = 1
a2 = 12! =
12
a3 = 13! =
11·2·3 < 1
22
an = 1n! =
11·2·3·····n < 1
2n−1
Hence Sn < 1 + 12 + 1
22 + · · ·+ 12n−1 =
1−( 12)
n
1− 12
= 2(1− 1
2n
)< 2
Hence {Sn}∞n=1 is bounded above.
Since an = 1n! > 0, for all n ∈ N and {Sn}∞n=1 is convergent,
∞∑n=1
1n! is
cgt.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 16 / 22
Positive Terms Series
Example:∞∑
n=1
1n!
Let an = 1n! , for all n ∈ N and Sn be the nth partial sum of the series.
a1 = 11! = 1
a2 = 12! =
12
a3 = 13! =
11·2·3 < 1
22
an = 1n! =
11·2·3·····n < 1
2n−1
Hence Sn < 1 + 12 + 1
22 + · · ·+ 12n−1 =
1−( 12)
n
1− 12
= 2(1− 1
2n
)< 2
Hence {Sn}∞n=1 is bounded above.
Since an = 1n! > 0, for all n ∈ N and {Sn}∞n=1 is convergent,
∞∑n=1
1n! is
cgt.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 16 / 22
Positive Terms Series
Example:∞∑
n=1
1n!
Let an = 1n! , for all n ∈ N and Sn be the nth partial sum of the series.
a1 = 11! = 1
a2 = 12! =
12
a3 = 13! =
11·2·3 < 1
22
an = 1n! =
11·2·3·····n < 1
2n−1
Hence Sn < 1 + 12 + 1
22 + · · ·+ 12n−1 =
1−( 12)
n
1− 12
= 2(1− 1
2n
)< 2
Hence {Sn}∞n=1 is bounded above.
Since an = 1n! > 0, for all n ∈ N and {Sn}∞n=1 is convergent,
∞∑n=1
1n! is
cgt.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 16 / 22
Positive Terms Series
Example:∞∑
n=1
1n!
Let an = 1n! , for all n ∈ N and Sn be the nth partial sum of the series.
a1 = 11! = 1
a2 = 12! =
12
a3 = 13! =
11·2·3 < 1
22
an = 1n! =
11·2·3·····n < 1
2n−1
Hence Sn < 1 + 12 + 1
22 + · · ·+ 12n−1 =
1−( 12)
n
1− 12
= 2(1− 1
2n
)< 2
Hence {Sn}∞n=1 is bounded above.
Since an = 1n! > 0, for all n ∈ N and {Sn}∞n=1 is convergent,
∞∑n=1
1n! is
cgt.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 16 / 22
Positive Terms Series
Theorem∞∑
n=1
1np is convergent if and only if p > 1.
Proof:(⇐)
Suppose p > 1. Let Sn =n∑
r=1
1rp , for all n ∈ N.
S2n−1 = 11p +
( 12p + 1
3p
)+( 1
4p + 15p + 1
6p + 17p
)+( 1
8p + · · ·+ 115p
)+ · · ·+(
1(2n−1)
p + · · ·+ 1(2n−1)p
)
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 17 / 22
Positive Terms Series
Theorem∞∑
n=1
1np is convergent if and only if p > 1.
Proof:(⇐)
Suppose p > 1. Let Sn =n∑
r=1
1rp , for all n ∈ N.
S2n−1 = 11p +
( 12p + 1
3p
)+( 1
4p + 15p + 1
6p + 17p
)+( 1
8p + · · ·+ 115p
)+ · · ·+(
1(2n−1)
p + · · ·+ 1(2n−1)p
)
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 17 / 22
Positive Terms Series
Theorem∞∑
n=1
1np is convergent if and only if p > 1.
Proof:(⇐)
Suppose p > 1. Let Sn =n∑
r=1
1rp , for all n ∈ N.
S2n−1 = 11p +
( 12p + 1
3p
)+( 1
4p + 15p + 1
6p + 17p
)+( 1
8p + · · ·+ 115p
)+ · · ·+(
1(2n−1)
p + · · ·+ 1(2n−1)p
)
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 17 / 22
Positive Terms Series
Theorem∞∑
n=1
1np is convergent if and only if p > 1.
Proof:(⇐)
Suppose p > 1. Let Sn =n∑
r=1
1rp , for all n ∈ N.
S2n−1 = 11p +
( 12p + 1
3p
)+( 1
4p + 15p + 1
6p + 17p
)+( 1
8p + · · ·+ 115p
)+ · · ·+(
1(2n−1)
p + · · ·+ 1(2n−1)p
)
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 17 / 22
Positive Terms Series
Theorem∞∑
n=1
1np is convergent if and only if p > 1.
Proof:(⇐)
Suppose p > 1. Let Sn =n∑
r=1
1rp , for all n ∈ N.
S2n−1 = 11p +
( 12p + 1
3p
)+( 1
4p + 15p + 1
6p + 17p
)+( 1
8p + · · ·+ 115p
)+ · · ·+(
1(2n−1)
p + · · ·+ 1(2n−1)p
)
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 17 / 22
Positive Terms Series
Proof Contd.
S2n−1 = 11p +
( 12p + 1
3p
)+( 1
4p + 15p + 1
6p + 17p
)+( 1
8p + · · ·+ 115p
)+ · · ·+(
1(2n−1)
p + · · ·+ 1(2n−1)p
)< 1
1p +( 1
2p + 12p
)+( 1
4p + 14p + 1
4p + 14p
)+( 1
8p + · · · 18p
)+ ...+(
1(2n−1)
p + · · ·+ 1(2n−1)
p
)= 1
1p + 2 · 12p + 4 · 1
4p + 8 · 18p + · · ·+ 2n−1 · 1
(2n−1)p
= 11p + 2 · 1
2p + 22 · 1(2p)2 + 23 · 1
(2p)3 + · · ·+ 2n−1 · 1(2p)n−1
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 18 / 22
Positive Terms Series
Proof Contd.
S2n−1 = 11p +
( 12p + 1
3p
)+( 1
4p + 15p + 1
6p + 17p
)+( 1
8p + · · ·+ 115p
)+ · · ·+(
1(2n−1)
p + · · ·+ 1(2n−1)p
)< 1
1p +( 1
2p + 12p
)+( 1
4p + 14p + 1
4p + 14p
)+( 1
8p + · · · 18p
)+ ...+(
1(2n−1)
p + · · ·+ 1(2n−1)
p
)= 1
1p + 2 · 12p + 4 · 1
4p + 8 · 18p + · · ·+ 2n−1 · 1
(2n−1)p
= 11p + 2 · 1
2p + 22 · 1(2p)2 + 23 · 1
(2p)3 + · · ·+ 2n−1 · 1(2p)n−1
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 18 / 22
Positive Terms Series
Proof Contd.
S2n−1 = 11p +
( 12p + 1
3p
)+( 1
4p + 15p + 1
6p + 17p
)+( 1
8p + · · ·+ 115p
)+ · · ·+(
1(2n−1)
p + · · ·+ 1(2n−1)p
)< 1
1p +( 1
2p + 12p
)+( 1
4p + 14p + 1
4p + 14p
)+( 1
8p + · · · 18p
)+ ...+(
1(2n−1)
p + · · ·+ 1(2n−1)
p
)= 1
1p + 2 · 12p + 4 · 1
4p + 8 · 18p + · · ·+ 2n−1 · 1
(2n−1)p
= 11p + 2 · 1
2p + 22 · 1(2p)2 + 23 · 1
(2p)3 + · · ·+ 2n−1 · 1(2p)n−1
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 18 / 22
Positive Terms Series
Proof Contd.
S2n−1 = 11p +
( 12p + 1
3p
)+( 1
4p + 15p + 1
6p + 17p
)+( 1
8p + · · ·+ 115p
)+ · · ·+(
1(2n−1)
p + · · ·+ 1(2n−1)p
)< 1
1p +( 1
2p + 12p
)+( 1
4p + 14p + 1
4p + 14p
)+( 1
8p + · · · 18p
)+ ...+(
1(2n−1)
p + · · ·+ 1(2n−1)
p
)= 1
1p + 2 · 12p + 4 · 1
4p + 8 · 18p + · · ·+ 2n−1 · 1
(2n−1)p
= 11p + 2 · 1
2p + 22 · 1(2p)2 + 23 · 1
(2p)3 + · · ·+ 2n−1 · 1(2p)n−1
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 18 / 22
Positive Terms Series
Proof Contd.
S2n−1 = 11p +
( 12p + 1
3p
)+( 1
4p + 15p + 1
6p + 17p
)+( 1
8p + · · ·+ 115p
)+ · · ·+(
1(2n−1)
p + · · ·+ 1(2n−1)p
)< 1
1p +( 1
2p + 12p
)+( 1
4p + 14p + 1
4p + 14p
)+( 1
8p + · · · 18p
)+ ...+(
1(2n−1)
p + · · ·+ 1(2n−1)
p
)= 1
1p + 2 · 12p + 4 · 1
4p + 8 · 18p + · · ·+ 2n−1 · 1
(2n−1)p
= 11p + 2 · 1
2p + 22 · 1(2p)2 + 23 · 1
(2p)3 + · · ·+ 2n−1 · 1(2p)n−1
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 18 / 22
Positive Terms Series
Proof Contd.
= 1 + 12p−1 + 1
(2p−1)2 + 1
(2p−1)3 + · · ·+ 1
(2p−1)n−1
=1−
(1
2p−1
)n
1− 12p−1
< 11− 1
2p−1, for all n ∈ N.
Since n ≤ 2n − 1, for all n ∈ N and {Sn}∞n=1 is an increasing sequencewe have,
0 < Sn ≤ S2n−1.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 19 / 22
Positive Terms Series
Proof Contd.
= 1 + 12p−1 + 1
(2p−1)2 + 1
(2p−1)3 + · · ·+ 1
(2p−1)n−1
=1−
(1
2p−1
)n
1− 12p−1
< 11− 1
2p−1, for all n ∈ N.
Since n ≤ 2n − 1, for all n ∈ N and {Sn}∞n=1 is an increasing sequencewe have,
0 < Sn ≤ S2n−1.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 19 / 22
Positive Terms Series
Proof Contd.
= 1 + 12p−1 + 1
(2p−1)2 + 1
(2p−1)3 + · · ·+ 1
(2p−1)n−1
=1−
(1
2p−1
)n
1− 12p−1
< 11− 1
2p−1, for all n ∈ N.
Since n ≤ 2n − 1, for all n ∈ N and {Sn}∞n=1 is an increasing sequencewe have,
0 < Sn ≤ S2n−1.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 19 / 22
Positive Terms Series
Proof Contd.
= 1 + 12p−1 + 1
(2p−1)2 + 1
(2p−1)3 + · · ·+ 1
(2p−1)n−1
=1−
(1
2p−1
)n
1− 12p−1
< 11− 1
2p−1, for all n ∈ N.
Since n ≤ 2n − 1, for all n ∈ N and {Sn}∞n=1 is an increasing sequencewe have,
0 < Sn ≤ S2n−1.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 19 / 22
Positive Terms Series
Proof Contd.
Therefore Sn ≤ 11− 1
2p−1for all n ∈ N
Hence , {Sn}∞n=1 is bounded above.
Thus,∞∑
n=1
1np is convergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 20 / 22
Positive Terms Series
Proof Contd.
Therefore Sn ≤ 11− 1
2p−1for all n ∈ N
Hence , {Sn}∞n=1 is bounded above.
Thus,∞∑
n=1
1np is convergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 20 / 22
Positive Terms Series
Proof Contd.
Therefore Sn ≤ 11− 1
2p−1for all n ∈ N
Hence , {Sn}∞n=1 is bounded above.
Thus,∞∑
n=1
1np is convergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 20 / 22
Positive Terms Series
Proof Contd.
Therefore Sn ≤ 11− 1
2p−1for all n ∈ N
Hence , {Sn}∞n=1 is bounded above.
Thus,∞∑
n=1
1np is convergent.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 20 / 22
Positive Terms Series
Proof Contd.
Now Suppose p ≤ 1,
S2n = 1 + 12p +
( 13p + 1
4p
)+( 1
5p + 16p + 1
7p + 18p
)+ · · ·+(
1(2n−1+1)
p + · · ·+ 1(2n)p
)
Since p ≤ 1, np ≤ n, for all n,⇒ 1np ≥ 1
n , for all n.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 21 / 22
Positive Terms Series
Proof Contd.
Now Suppose p ≤ 1,
S2n = 1 + 12p +
( 13p + 1
4p
)+( 1
5p + 16p + 1
7p + 18p
)+ · · ·+(
1(2n−1+1)
p + · · ·+ 1(2n)p
)
Since p ≤ 1, np ≤ n, for all n,⇒ 1np ≥ 1
n , for all n.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 21 / 22
Positive Terms Series
Proof Contd.
Now Suppose p ≤ 1,
S2n = 1 + 12p +
( 13p + 1
4p
)+( 1
5p + 16p + 1
7p + 18p
)+ · · ·+(
1(2n−1+1)
p + · · ·+ 1(2n)p
)
Since p ≤ 1, np ≤ n, for all n,⇒ 1np ≥ 1
n , for all n.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 21 / 22
Positive Terms Series
Proof Contd.
Now Suppose p ≤ 1,
S2n = 1 + 12p +
( 13p + 1
4p
)+( 1
5p + 16p + 1
7p + 18p
)+ · · ·+(
1(2n−1+1)
p + · · ·+ 1(2n)p
)
Since p ≤ 1, np ≤ n, for all n,⇒ 1np ≥ 1
n , for all n.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 21 / 22
Positive Terms Series
Proof Contd.
Hence,
S2n ≥(
1 +12
)+
(14+
14
)+
(18+ · · ·+ 1
8
)+ · · ·+
(12n + · · ·+ 1
2n
)= 1 +
12+ 2 · 1
4+ 4 · 1
8+ · · ·+ 2n−1 · 1
2n
= 1 +12+
12+
12+ · · ·+ 1
2= 1 +
n2, for all n.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence,∞∑
n=1
1np is divergent, if p ≤ 1.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 22 / 22
Positive Terms Series
Proof Contd.
Hence,
S2n ≥(
1 +12
)+
(14+
14
)+
(18+ · · ·+ 1
8
)+ · · ·+
(12n + · · ·+ 1
2n
)= 1 +
12+ 2 · 1
4+ 4 · 1
8+ · · ·+ 2n−1 · 1
2n
= 1 +12+
12+
12+ · · ·+ 1
2= 1 +
n2, for all n.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence,∞∑
n=1
1np is divergent, if p ≤ 1.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 22 / 22
Positive Terms Series
Proof Contd.
Hence,
S2n ≥(
1 +12
)+
(14+
14
)+
(18+ · · ·+ 1
8
)+ · · ·+
(12n + · · ·+ 1
2n
)= 1 +
12+ 2 · 1
4+ 4 · 1
8+ · · ·+ 2n−1 · 1
2n
= 1 +12+
12+
12+ · · ·+ 1
2= 1 +
n2, for all n.
Since, S2n →∞ as n→∞, {Sn}∞n=1 is divergent.
Hence,∞∑
n=1
1np is divergent, if p ≤ 1.
Dr. G.H.J. Lanel (USJP) MAT 127 2.0 Calculus II Lecture 2 22 / 22
