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MATH 38Mathematical Analysis III
I. F. Evidente
IMSP (UPLB)
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Previously...
1 Convergence of an infinite series by definition2 Divergence Test3 Tests for Infinite Series of Nonnegative Terms:
1 Integral Test2 Limit Comparison Test
4 Special Series: geometric, harmonic, constant, p-series
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Outline
1 Tests for Series of Nonnegative Terms (continued)Comparison Test
2 Test for Alternating Series
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Theorem (Two of the Four Convergence Theorems)Suppose
an and
bn are infinite series and c is a nonzero constant.
1 Ifan converges, then
can converges and
can = can . If an
diverges, thencan diverges.
2 Ifan and
bn differ only by a finite number of terms, then either
both converge or both diverge.
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Example
1n=k
1
n, where k is any integer greater than 1
(divergent)
2n=k
5
n7/5(convergent)
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Example
1n=k
1
n, where k is any integer greater than 1 (divergent)
2n=k
5
n7/5(convergent)
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Example
1n=k
1
n, where k is any integer greater than 1 (divergent)
2n=k
5
n7/5
(convergent)
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Example
1n=k
1
n, where k is any integer greater than 1 (divergent)
2n=k
5
n7/5(convergent)
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Outline
1 Tests for Series of Nonnegative Terms (continued)Comparison Test
2 Test for Alternating Series
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Theorem (Comparison Test)Letan and
bn be infinite series of nonnegative terms such that an bn
for all n = 1,2, ...
1 Ifbn converges, then
an converges.
2 Ifan diverges, then
bn diverges.
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Theorem (Comparison Test)Letan and
bn be infinite series of nonnegative terms such that an bn
for all n = 1,2, ...1 If
bn converges, then
an converges.
2 Ifan diverges, then
bn diverges.
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Theorem (Comparison Test)Letan and
bn be infinite series of nonnegative terms such that an bn
for all n = 1,2, ...1 If
bn converges, then
an converges.
2 Ifan diverges, then
bn diverges.
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RemarkDifficulty with comparison test: you have to make an intelligent guess as towhether or not the series is convergent.
If your guess is convergent, try to find a "greater" convergent specialseries.If your guess divergent, try to find a "lesser" divergent special series.
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RemarkDifficulty with comparison test: you have to make an intelligent guess as towhether or not the series is convergent.
If your guess is convergent,
try to find a "greater" convergent specialseries.If your guess divergent, try to find a "lesser" divergent special series.
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RemarkDifficulty with comparison test: you have to make an intelligent guess as towhether or not the series is convergent.
If your guess is convergent, try to find a "greater" convergent specialseries.
If your guess divergent, try to find a "lesser" divergent special series.
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RemarkDifficulty with comparison test: you have to make an intelligent guess as towhether or not the series is convergent.
If your guess is convergent, try to find a "greater" convergent specialseries.If your guess divergent,
try to find a "lesser" divergent special series.
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RemarkDifficulty with comparison test: you have to make an intelligent guess as towhether or not the series is convergent.
If your guess is convergent, try to find a "greater" convergent specialseries.If your guess divergent, try to find a "lesser" divergent special series.
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With some abuse of notation :)
TipConvergent:
an
bn
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With some abuse of notation :)
TipConvergent:
an
bn
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With some abuse of notation :)
TipConvergent:
an
bn
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With some abuse of notation :)
TipConvergent:
an
bn
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With some abuse of notation :)
TipConvergent:
an
bn
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TipThe comparison test could be useful if f (n) involves sin, cos, ln or Arctanbecause:
1 sinn 1; 1 cosn 10 lnn pn n if n 1pi2
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y = x
1
y =pxy = lnx
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Remark
1 If 0< a < b, then 1a> 1
b
2 Given a positive fraction: If the numerator is held constant and1 the denominator is increased, the fraction will decrease2 the denominator is decreased, the fraction will increase
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Remark
1 If 0< a < b, then 1a> 1
b2 Given a positive fraction: If the numerator is held constant and
1 the denominator is increased, the fraction will decrease2 the denominator is decreased, the fraction will increase
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Examplen=1
5
n3n
n=1
5
n3n
n=1
5
3n
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Examplen=1
5
n3n
n=1
5
n3n
n=1
5
3n
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Examplen=1
5
n3n
n=1
5
n3n
n=1
5
3n
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Examplen=1
5
n3n
n=1
5
n3n
n=1
5
3n
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Examplen=1
|cosn|n2+1
n=1
|cosn|n2+1
n=1
1
n2
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Examplen=1
|cosn|n2+1
n=1
|cosn|n2+1
n=1
1
n2
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Examplen=1
|cosn|n2+1
n=1
|cosn|n2+1
n=1
1
n2
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Examplen=1
|cosn|n2+1
n=1
|cosn|n2+1
n=1
1
n2
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Examplen=2
1
lnn
n=2
1
lnn>
n=2
1
n>
The series is divergent.
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Examplen=2
1
lnn
n=2
1
lnn>
n=2
1
n
>
The series is divergent.
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Examplen=2
1
lnn
n=2
1
lnn>
n=2
1
n>
The series is divergent.
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Examplen=2
1
lnn
n=2
1
lnn>
n=2
1
n>
The series is divergent.
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Examplen=1
(1+ 2
n
)n
n=1
(1+ 2
n
)n>
n=1
1>
The series is divergent.
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Examplen=1
(1+ 2
n
)n
n=1
(1+ 2
n
)n>
n=1
1
>
The series is divergent.
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Examplen=1
(1+ 2
n
)n
n=1
(1+ 2
n
)n>
n=1
1>
The series is divergent.
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Examplen=1
(1+ 2
n
)n
n=1
(1+ 2
n
)n>
n=1
1>
The series is divergent.
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Outline
1 Tests for Series of Nonnegative Terms (continued)Comparison Test
2 Test for Alternating Series
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DefinitionAn alternating series is a series whose terms form an alternatingsequence.
(That is, the signs alternate from positive to negative.)
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DefinitionAn alternating series is a series whose terms form an alternatingsequence. (That is, the signs alternate from positive to negative.)
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Theorem (Alternating Series Test for Convergence)Letan be an alternating series.
The alternating series is convergent ifboth of the following conditions hold:
1 limn |an | = 0
2 {|an |} is a decreasing sequence. (Recall two ways to show a sequenceis decreasing.)
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Theorem (Alternating Series Test for Convergence)Letan be an alternating series. The alternating series is convergent if
both of the following conditions hold:
1 limn |an | = 0
2 {|an |} is a decreasing sequence. (Recall two ways to show a sequenceis decreasing.)
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Theorem (Alternating Series Test for Convergence)Letan be an alternating series. The alternating series is convergent if
both of the following conditions hold:1 lim
n |an | = 0
2 {|an |} is a decreasing sequence. (Recall two ways to show a sequenceis decreasing.)
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Theorem (Alternating Series Test for Convergence)Letan be an alternating series. The alternating series is convergent if
both of the following conditions hold:1 lim
n |an | = 02 {|an |} is a decreasing sequence.
(Recall two ways to show a sequenceis decreasing.)
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Theorem (Alternating Series Test for Convergence)Letan be an alternating series. The alternating series is convergent if
both of the following conditions hold:1 lim
n |an | = 02 {|an |} is a decreasing sequence. (Recall two ways to show a sequence
is decreasing.)
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n1 lim
n1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an |
= (1)nn
= 1n1 lim
n1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n
1 limn
1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n
1 limn
1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n1 lim
n1
n
= 02 Is {|an |} decreasing?
Let f (n)= 1n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n1 lim
n1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n1 lim
n1
n= 0
2 Is {|an |} decreasing?
Let f (n)= 1n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n1 lim
n1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n1 lim
n1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)
= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n1 lim
n1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n1 lim
n1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n1 lim
n1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)= 1n2
< 0
Yes, {|an |} decreasing.
The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n1 lim
n1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)nn
(Alternating Harmonic Series)
|an | = (1)nn
= 1n1 lim
n1
n= 0
2 Is {|an |} decreasing?Let f (n)= 1
n
f (n)= 1n2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
RemarkThe alternating harmonic series is convergent.
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Examplen=1
(1)n1lnn
|an |
= (1)n1lnn
= 1lnn1 lim
n1
lnn= 0
2 Is {|an |} decreasing?Let f (n)= 1
lnn
f (n)= 1n(lnn)2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
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Examplen=1
(1)n1lnn
|an | = (1)n1lnn
= 1lnn
1 limn
1
lnn= 0
2 Is {|an |} decreasing?Let f (n)= 1
lnn
f (n)= 1n(lnn)2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
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Examplen=1
(1)n1lnn
|an | = (1)n1lnn
= 1lnn
1 limn
1
lnn= 0
2 Is {|an |} decreasing?Let f (n)= 1
lnn
f (n)= 1n(lnn)2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
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Examplen=1
(1)n1lnn
|an | = (1)n1lnn
= 1lnn1 lim
n1
lnn
= 02 Is {|an |} decreasing?
Let f (n)= 1lnn
f (n)= 1n(lnn)2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
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Examplen=1
(1)n1lnn
|an | = (1)n1lnn
= 1lnn1 lim
n1
lnn= 0
2 Is {|an |} decreasing?Let f (n)= 1
lnn
f (n)= 1n(lnn)2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
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Examplen=1
(1)n1lnn
|an | = (1)n1lnn
= 1lnn1 lim
n1
lnn= 0
2 Is {|an |} decreasing?
Let f (n)= 1lnn
f (n)= 1n(lnn)2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
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Examplen=1
(1)n1lnn
|an | = (1)n1lnn
= 1lnn1 lim
n1
lnn= 0
2 Is {|an |} decreasing?Let f (n)= 1
lnn
f (n)= 1n(lnn)2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
-
Examplen=1
(1)n1lnn
|an | = (1)n1lnn
= 1lnn1 lim
n1
lnn= 0
2 Is {|an |} decreasing?Let f (n)= 1
lnn
f (n)
= 1n(lnn)2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
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Examplen=1
(1)n1lnn
|an | = (1)n1lnn
= 1lnn1 lim
n1
lnn= 0
2 Is {|an |} decreasing?Let f (n)= 1
lnn
f (n)= 1n(lnn)2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
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Examplen=1
(1)n1lnn
|an | = (1)n1lnn
= 1lnn1 lim
n1
lnn= 0
2 Is {|an |} decreasing?Let f (n)= 1
lnn
f (n)= 1n(lnn)2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
-
Examplen=1
(1)n1lnn
|an | = (1)n1lnn
= 1lnn1 lim
n1
lnn= 0
2 Is {|an |} decreasing?Let f (n)= 1
lnn
f (n)= 1n(lnn)2
< 0
Yes, {|an |} decreasing.
The series is convergent since it passes AST.
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Examplen=1
(1)n1lnn
|an | = (1)n1lnn
= 1lnn1 lim
n1
lnn= 0
2 Is {|an |} decreasing?Let f (n)= 1
lnn
f (n)= 1n(lnn)2
< 0
Yes, {|an |} decreasing. The series is convergent since it passes AST.
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Remark1 Please use this only for an alternating series. :)
2 If an alternating series does not satisfy the above conditions, youcannot conclude that the alternating series is divergent.
3 If the first condition is not satisfied, use the divergence test to provethat the sequence is divergent.
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Remark1 Please use this only for an alternating series. :)2 If an alternating series does not satisfy the above conditions, you
cannot conclude that the alternating series is divergent.
3 If the first condition is not satisfied, use the divergence test to provethat the sequence is divergent.
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Remark1 Please use this only for an alternating series. :)2 If an alternating series does not satisfy the above conditions, you
cannot conclude that the alternating series is divergent.3 If the first condition is not satisfied, use the divergence test to prove
that the sequence is divergent.
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Examplen=1
(1)nnn+1
|an | = (1)nnn+1
= nn+1limn
n
n+1 = 1 limnn
n+1 =1
This implies: limn
(1)nnn+1 does not exist (see remark after Thm 1.1.3).
The series is divergent by the Divergence Test.
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Examplen=1
(1)nnn+1
|an |
= (1)nnn+1
= nn+1limn
n
n+1 = 1 limnn
n+1 =1
This implies: limn
(1)nnn+1 does not exist (see remark after Thm 1.1.3).
The series is divergent by the Divergence Test.
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Examplen=1
(1)nnn+1
|an | = (1)nnn+1
= nn+1
limn
n
n+1 = 1 limnn
n+1 =1
This implies: limn
(1)nnn+1 does not exist (see remark after Thm 1.1.3).
The series is divergent by the Divergence Test.
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Examplen=1
(1)nnn+1
|an | = (1)nnn+1
= nn+1
limn
n
n+1 = 1 limnn
n+1 =1
This implies: limn
(1)nnn+1 does not exist (see remark after Thm 1.1.3).
The series is divergent by the Divergence Test.
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Examplen=1
(1)nnn+1
|an | = (1)nnn+1
= nn+1limn
n
n+1
= 1 limn
n
n+1 =1
This implies: limn
(1)nnn+1 does not exist (see remark after Thm 1.1.3).
The series is divergent by the Divergence Test.
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Examplen=1
(1)nnn+1
|an | = (1)nnn+1
= nn+1limn
n
n+1 = 1
limn
n
n+1 =1
This implies: limn
(1)nnn+1 does not exist (see remark after Thm 1.1.3).
The series is divergent by the Divergence Test.
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Examplen=1
(1)nnn+1
|an | = (1)nnn+1
= nn+1limn
n
n+1 = 1 limnn
n+1
=1
This implies: limn
(1)nnn+1 does not exist (see remark after Thm 1.1.3).
The series is divergent by the Divergence Test.
-
Examplen=1
(1)nnn+1
|an | = (1)nnn+1
= nn+1limn
n
n+1 = 1 limnn
n+1 =1
This implies: limn
(1)nnn+1 does not exist (see remark after Thm 1.1.3).
The series is divergent by the Divergence Test.
-
Examplen=1
(1)nnn+1
|an | = (1)nnn+1
= nn+1limn
n
n+1 = 1 limnn
n+1 =1
This implies: limn
(1)nnn+1
does not exist (see remark after Thm 1.1.3). The series is divergent by the Divergence Test.
-
Examplen=1
(1)nnn+1
|an | = (1)nnn+1
= nn+1limn
n
n+1 = 1 limnn
n+1 =1
This implies: limn
(1)nnn+1 does not exist (see remark after Thm 1.1.3).
The series is divergent by the Divergence Test.
-
Examplen=1
(1)nnn+1
|an | = (1)nnn+1
= nn+1limn
n
n+1 = 1 limnn
n+1 =1
This implies: limn
(1)nnn+1 does not exist (see remark after Thm 1.1.3).
The series is divergent by the Divergence Test.
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The End. Thank you.
Tests for Series of Nonnegative Terms (continued)Comparison Test
Test for Alternating Series