Calculus Of One Real Variable By Pheng Kim Ving


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1. Alternating Series 
Alternating
Series
A series may have finitely many negative terms, and thus it
must have infinitely many positiveor0valued terms; they're
ultimately positive series. A series may have infinitely many negative terms,
and thus it may have either no or finitely or
infinitely many positiveor0valued terms. All the tests discussed in previous
sections apply only to ultimately positive series. If
a series has infinitely many negative terms and no or finitely many positive
terms, the tests can be applied to the series of the
opposite or negative of its terms, which is an (ultimately) positive series,
and which converges iff the original series converges.
If a series has infinitely many negative terms and
infinitely many positive terms, no previous test can be applied to it directly.
In
this section we present two tests: the alternatingseries test, which applies
to alternating series, and the absoluteconvergence
test, which applies to any kind of series.
The series:
Definition 1.1
Alternating Series
A series is said to be an alternating series if its terms alternate in sign. 
The
AlternatingSeries Test (AST)
The harmonic series:
The second one is just the negative of the first one. One
converges iff the other converges. Here we deal with the first one.
Does it diverge to infinity too, or does it diverge to negative infinity, or
does it simply diverge, or does it converge? Refer to Fig.
1.1. Let a_{n} and s_{n} be the nth term and nth partial
sum respectively. Note that a_{n}
= 1/n > 0 if n
is odd and a_{n} = 1/n < 0 if n is

Fig. 1.1 Partial Sums Of 
even. We have:
etc. If n is odd,
then a_{n}_{+1} = (1)^{(}^{n}^{+1)}^{+1}/n = 1/n < 0, and
s_{n} jumps downward to s_{n}_{+1} = s_{n} + a_{n}_{+1} by an amount
equal to the size
of the next term, namely a_{n}_{+1}. If n is even, then a_{n}_{+1} = 1/n > 0, and s_{n}
jumps upward to s_{n}_{+1} by an amount
equal to the next
term, namely a_{n}_{+1}. Since a_{n} approaches 0 as n approaches infinity, the size of the
oscillation of s_{n} as
a function of n is
decreasing toward 0. In fact for any positive integer n
we get:
s_{2} < s_{4} < s_{6} < s_{8} < s_{10} < ... < s_{2}_{n}_{2} < s_{2}_{n} < s_{2}_{n}_{1} < s_{2}_{n}_{3} < ... < s_{9} < s_{7} < s_{5} < s_{3} < s_{1}.
Each evennumbered partial sum is less than every
oddnumbered partial sum. The sequence of evennumbered partial sums
is increasing. The sequence of oddnumbered partial sums is decreasing. As the
size of the oscillation back and forth between
the two sequences approaches 0, the gap between them approaches 0 as n approaches infinity, and thus they converge to
the
same number. Consequently the sequence of partial sums and hence the
alternating harmonic series converge to that number.
The following test asserts that a series converges provided
it's an alternating series and satisfies two more conditions. It's
called the alternatingseries test, abbreviated as AST.
Theorem 1.1 The
AlternatingSeries Test (AST)
then it converges. 
Proof
Suppose a_{1} > 0. Then a_{n}
> 0 if n is odd and a_{n}
< 0 if n is even. Let s_{n}
be the nth partial sum. For any n > 0 we have:
The proof for the case where a_{1} < 0 is
similar.
EOP
The Ultimately
Version
Theorem 1.1 is also valid for series that's ultimately
alternating and/or where the sequence of the absolute values of its terms
is ultimately decreasing.
The
Ultimately Version Of The AlternatingSeries Test
then it converges. 
Proof
EOP
Remarks 1.1
1. For the decreasing condition, we have to consider
the absolute values, since the terms themselves are neither decreasing
nor increasing.
2. A sequence that's decreasing in absolute values
doesn't necessarily approach 0. For example, the alternating harmonic
sequence has the absolute values of
its terms decreasing and its terms approaching 0, while:
has its terms approaching 0 but their absolute values neither decreasing nor increasing.
Example 1.1
Show that the alternating harmonic series converges.
Solution
The sequence of the absolute values of its terms is decreasing and the sequence
of its terms approaches 0. So it converges.
EOS
Example 1.2
Discuss the behavior (convergence or divergence) of the following series:
Solution
a. The series is alternating. The absolute values of
its terms are decreasing. Its terms approach 0. So, by the AST, it
converges.
EOS
Example 1.3
Determine the convergence or divergence of this series:
Solution
The series is alternating. Let a_{n}
= (1)^{n}^{+1} ( ln
n)/n. We have:
converges, by the AST.
EOS
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2. Absolute Convergence 
Absolute
Convergence
Given a series, the series of the absolute values of its
terms is a positive series, to which the SCT, LCT, RooT, RaT, and IT can
be applied. This fact motivates the investigation of absolute convergence.
Consider the series:
Definition 2.1
Absolute Convergence
A series is said to converge absolutely, or to be absolutely
convergent, if the series of the absolute values of its terms 
The
AbsoluteConvergence Test (ACT)
For series that have infinitely many negative terms and
infinitely many positive terms and that aren't alternating, the tests
discussed in previous sections and the AST can't be applied. Suppose a series
converges absolutely. The sum of its terms is
less than or equal to the sum of the absolute values of the terms, which is a
(finite) number because the series converges
absolutely. So the sum of the terms can't be infinity or negative infinity. Can
the sum of the terms oscillate between two values
without the oscillation approaching 0 (eg, the sum oscillates between 100 and
102 as the terms oscillate between 2 and 2)?
No, because if it did, then the absolute values of the terms wouldn't approach
0 and the series wouldn't converge absolutely.
Thus the sum of the terms must be a (finite) number too. That is, the series
converges too.
For a series that has infinitely many negative terms and
infinitely many positive terms and that's not alternating, just like any
series, if it converges absolutely then it converges. In general, the
cancellation that occurs because some terms are positive
and others are negative makes it easier for a series to converge than if all
the terms are of one sign. The implication of the
convergence by the absolute convergence can be used to test the convergence of
such series. Testing the convergence of a
series by examining its absolute convergence is known as the absoluteconvergence
test, or ACT for short.
Theorem 2.1 The
AbsoluteConvergence Test (ACT)
If a series converges absolutely then it converges. 
Note On The Proof
original series = first defined series + second defined series.
The first defined series contains the positive terms and the
second one contains the negative terms of the original series. For
the first one, we replace every negative term in the original series by 0, and
for the second one, we replace every positive term
in the original series by 0. These replacements are made in order to have the
original series as the sum of the two defined
series.
Remarks 2.1
1. Again the ACT can be used to test the convergence
of a series that has infinitely many negative terms and infinitely many
positive terms and that's not
alternating.
4. Every convergent positive series is also absolutely convergent, for it's identical to its corresponding series of absolute values.
Testing For
Absolute Convergence
Example 2.1
Establish the behavior (convergence or divergence) of the series:
with the pattern that two positive terms are followed by one negative term, starting with two positive terms.
Note
That series has infinitely many negative terms and
infinitely many positive terms. So the SCT, LCT, RooT, RaT, IT, and pseries
property can't be applied to it or to its opposite (negative). It's not
alternating. Thus the AST can't be applied either. The
remaining hope is the ACT.
Solution
The series of the absolute values of the terms of the given series is:
which is a pseries with
p = 3 > 1 and thus converges. Since the given
series converges absolutely, it converges, by the ACT.
EOS
Conditional
Convergence
Definition 2.2
Conditional Convergence
A series is said to converge conditionally, or to
be conditionally convergent, if it converges but doesn't converge 
Remarks 2.2
1. As an example, the alternating harmonic series converges conditionally.
2. Conditional convergence and absolute convergence
are mutually exclusive properties. If a series is conditionally convergent,
then it's not absolutely convergent,
and if a series is absolutely convergent, then it's not conditionally convergent.
3. If a series doesn't converge absolutely, there
remains the possibility that it converges conditionally. So again, we can't
conclude that a
nonabsolutelyconvergent series is nonconvergent. We have to try other tests
to determine its convergence
or divergence.
4. If a series converges, then it converges either also absolutely or just conditionally.
5. If a series satisfies the AST then it converges.
It doesn't necessarily converge conditionally. It converges conditionally only
if
it doesn't converge absolutely.
Example 2.2
Solution
The corresponding series of absolute values is the harmonic series, which
diverges. So the given series can't be absolutely
convergent. Clearly, by the AST, the given series converges. Thus the given
series converges conditionally.
EOS
Example 2.3
Classify the series:
as absolutely convergent, conditionally convergent, or divergent.
Solution
Let a_{n} = (2)^{n}/n!. We then have:
EOS
As the two examples above show, when classifying a series as
absolutely convergent, conditionally convergent, or divergent,
we should first check for absolute convergence. If it's absolutely convergent,
we're done. If it's not, we check for convergence.
It's a waste of time to check for convergence first, because if it's
convergent, we still have to check for absolute convergence.
Example 2.4
Classify the series:
as absolutely convergent, conditionally convergent, or divergent.
Solution
EOS
Example 2.5
Classify this series:
as absolutely convergent, conditionally convergent, or divergent.
Solution
EOS
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3. Series Whose Terms Involve Functions Of A Real Variable 
Consider the series:
are members of the divergent family.
Example 3.1
For what values of x does the series:
converge absolutely? Converge conditionally? Diverge?
Solution
Let a_{n} = (x 3)^{n}/4^{n}n. We have:
converges absolutely for 1 < x
< 7, converges conditionally for x = 1, and
diverges everywhere else.
EOS
Example 3.2
Determine the values of x for which this series:
converges absolutely, converges conditionally, or diverges.
Solution
Let a_{n} = (n + 3)^{2} (x/(x + 2))^{n}. Then:
EOS
We can square both sides of x < x
+ 2 and retain
the direction of the inequality because absolute values are greater than
or equal to 0.
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4. Kinds Of Series And Tests 
The tests that have been discussed in previous sections and
in this section, in order of appearance, are the SCT, LCT, RooT,
RaT, IT, AST, and ACT. The pseries
property has also been discussed.
1. For an ultimately positive series (finite number of negative terms), all the tests except the AST can be applied.
2. For an ultimately negative series (finite number
of positive terms), all the tests except the AST can be applied to its
negative,
which is an ultimately positive
series.
3. For a series that's neither ultimately positive
nor ultimately negative (infinitely many negative terms and infinitely many
positive terms) nor alternating, the
ACT can be applied.
4. For an alternating series, the AST and ACT can be applied.
5. The ACT can be applied to any kind of series.
6. For a pseries, of course the pseries property can be applied.
Problems & Solutions 
1. Mark each of the following as T (true) or F
(false).
Solution
2.
a. The following statement is false. Give a counterexample.
Solution
a. The alternating harmonic series:
3. Consider the series:
a. Does the AST apply directly to it?
b. Classify it as absolutely convergent, conditionally convergent, or
divergent.
Solution
which obviously, by the AST, converges. Consequently the given series is conditionally convergent.
4. Classify the series:
as absolutely convergent, conditionally convergent, or divergent.
Solution
is convergent, by the AST, and consequently is conditionally convergent.
5. For what values of x does the
series:
converge absolutely? Converge conditionally? Diverge?
Solution
Let a_{n} = x^{n}/(2^{n} ln n). Then:
So the given series converges absolutely for 2 < x < 2, converges conditionally for x = 2, and diverges everywhere else.
6. See the Proof Of Theorem 2.1 for this problem. Let:
Solution
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