§12.4–Comparison Tests
Mark Woodard
Furman University
Fall 2010
Mark Woodard (Furman University)
§12.4–Comparison Tests
Fall 2010
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Outline
1
The p-series test
2
Comparison tests
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§12.4–Comparison Tests
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The p-series test
Definition
Let p > 0. A p-series is any series of the form
∞
X
1
.
kp
k=1
Mark Woodard (Furman University)
§12.4–Comparison Tests
Fall 2010
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The p-series test
Definition
Let p > 0. A p-series is any series of the form
∞
X
1
.
kp
k=1
Example
The harmonic series is a p-series with p = 1.
Mark Woodard (Furman University)
§12.4–Comparison Tests
Fall 2010
3 / 10
The p-series test
Definition
Let p > 0. A p-series is any series of the form
∞
X
1
.
kp
k=1
Example
The harmonic series is a p-series with p = 1.
Theorem (The p-Series Test (PST))
A p-series converges if p > 1 and diverges if 0 < p ≤ 1.
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§12.4–Comparison Tests
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Comparison tests
Comparison tests
P
The idea behindP
a comparison test is this: given a series
an construct a
reference series
bn in such a wayPthat the convergence or divergence of
P
an can be inferred from that of
bn .
Mark Woodard (Furman University)
§12.4–Comparison Tests
Fall 2010
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Comparison tests
Comparison tests
P
The idea behindP
a comparison test is this: given a series
an construct a
reference series
bn in such a wayPthat the convergence or divergence of
P
an can be inferred from that of
bn .
Theorem (Simple Comparison Test (SCT))
Suppose that 0 < an ≤ bn for n ≥ 1.
Mark Woodard (Furman University)
§12.4–Comparison Tests
Fall 2010
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Comparison tests
Comparison tests
P
The idea behindP
a comparison test is this: given a series
an construct a
reference series
bn in such a wayPthat the convergence or divergence of
P
an can be inferred from that of
bn .
Theorem (Simple Comparison Test (SCT))
Suppose that 0 < an ≤ bn for n ≥ 1.
P
P
If
bn converges, then
an converges.
Mark Woodard (Furman University)
§12.4–Comparison Tests
Fall 2010
4 / 10
Comparison tests
Comparison tests
P
The idea behindP
a comparison test is this: given a series
an construct a
reference series
bn in such a wayPthat the convergence or divergence of
P
an can be inferred from that of
bn .
Theorem (Simple Comparison Test (SCT))
Suppose that 0 < an ≤ bn for n ≥ 1.
P
P
If
bn converges, then
an converges.
P
P
If
an diverges, then
bn diverges.
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§12.4–Comparison Tests
Fall 2010
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Comparison tests
Proof.
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Comparison tests
Proof.
Let sn = a1 + · · · an and let tn = b1 + · · · + bn . Then we know that
sn ≤ tn for n ≥ 1.
Mark Woodard (Furman University)
§12.4–Comparison Tests
Fall 2010
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Comparison tests
Proof.
Let sn = a1 + · · · an and let tn = b1 + · · · + bn . Then we know that
sn ≤ tn for n ≥ 1.
P
If
bn converges, then tn is bounded above and, hence, so is sn ;
thus, sn converges.
Mark Woodard (Furman University)
§12.4–Comparison Tests
Fall 2010
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Comparison tests
Proof.
Let sn = a1 + · · · an and let tn = b1 + · · · + bn . Then we know that
sn ≤ tn for n ≥ 1.
P
If
bn converges, then tn is bounded above and, hence, so is sn ;
thus, sn converges.
P
If
an diverges, then sn is not bounded above and, hence, tn is not
bounded above either; thus, tn diverges.
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§12.4–Comparison Tests
Fall 2010
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Comparison tests
Problem
Determine the convergence or divergence of the following series:
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§12.4–Comparison Tests
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Comparison tests
Problem
Determine the convergence or divergence of the following series:
∞
X
ln(n + 2)
n=1
n
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§12.4–Comparison Tests
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Comparison tests
Problem
Determine the convergence or divergence of the following series:
∞
X
ln(n + 2)
n
n=1
∞
X
n=1
n2
1
+9
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§12.4–Comparison Tests
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Comparison tests
Problem
Determine the convergence or divergence of the following series:
∞
X
ln(n + 2)
n
n=1
∞
X
n=1
n2
1
+9
2n
1
−1
∞
X
n=1
Mark Woodard (Furman University)
§12.4–Comparison Tests
Fall 2010
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Comparison tests
Problem
Determine the convergence or divergence of the following series:
∞
X
ln(n + 2)
n
n=1
∞
X
n=1
n2
1
+9
2n
1
−1
∞
X
n=1
∞
X
2 − sin(n)
n=1
n
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§12.4–Comparison Tests
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Comparison tests
Theorem (The Limit Comparison Test (LCT))
If {an } and {bn } are sequences
P of positive
P numbers with an /bn → c and if
0 < c < ∞, then the series
an and
bn converge or diverge together.
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Comparison tests
Proof.
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Comparison tests
Proof.
There exists N such that |an /bn − c| < c/2 for n ≥ N.
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§12.4–Comparison Tests
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Comparison tests
Proof.
There exists N such that |an /bn − c| < c/2 for n ≥ N.
In other words, for n ≥ N,
c
3c
· bn ≤ an ≤
· bn
2
2
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§12.4–Comparison Tests
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Comparison tests
Proof.
There exists N such that |an /bn − c| < c/2 for n ≥ N.
In other words, for n ≥ N,
c
3c
· bn ≤ an ≤
· bn
2
2
Now the result follows from the SCT.
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§12.4–Comparison Tests
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Comparison tests
Problem
Determine the convergence or divergence of the following series:
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§12.4–Comparison Tests
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Comparison tests
Problem
Determine the convergence or divergence of the following series:
∞
X
1
n
2 −1
n=1
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§12.4–Comparison Tests
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Comparison tests
Problem
Determine the convergence or divergence of the following series:
∞
X
1
n
2 −1
n=1
∞
X
2n + 4
√
n4 + 2
n=1
Mark Woodard (Furman University)
§12.4–Comparison Tests
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Comparison tests
Solution
We compare with 1/2n :
1/(2n − 1)
1
2n
=
=
1/2n
2n − 1
1 − 2−n
P
1
= 1. The comparison is valid. Since
1/2n
−n
1−2
P
converges (being geometric),
1/(2n − 1) converges by LCT.
and lim
n→∞
Mark Woodard (Furman University)
§12.4–Comparison Tests
Fall 2010
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Comparison tests
Solution
We compare with 1/2n :
1/(2n − 1)
1
2n
=
=
1/2n
2n − 1
1 − 2−n
P
1
= 1. The comparison is valid. Since
1/2n
−n
1−2
P
converges (being geometric),
1/(2n − 1) converges by LCT.
and lim
n→∞
We compare with 1/n:
√
2 + n4
(2n + 4)/ n4 + 2
2n2 + 4n
= √
=q
1/n
n4 + 2
1 + n24
P
2 + n4
and lim q
= 2. The comparison is valid. Since
1/n
n→∞
1 + n24
√
P
diverges (PST, p = 1), (2n + 4)/ n4 + 2 diverges by LCT.
Mark Woodard (Furman University)
§12.4–Comparison Tests
Fall 2010
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