Calculus Of One Real Variable – By Pheng Kim Ving


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1. The Standard Comparison Test (SCT) 
In Section
14.2 we introduced some series that were shown to converge or diverge. To
determine the behavior (convergence or
divergence) of most of them, we algebraically manipulated the sequence of their
partial sums, expressing the general nth
partial sum as an explicit function of n and
evaluated its limit as n approaches
infinity. But for most other series, this approach
doesn't work, and thus the exact sums of those series usually can't be
determined. However there are techniques to determine
whether a given series converges, and, if it does, to approximate its sum to
any desired degree of accuracy. In this and the
next three sections we'll discuss some of these techniques. In this section
we'll compare series whose behavior is unknown to
series whose behavior is known. This requires that we have a stockpile of
series that are known to converge or diverge. For
convenience we display some such series below, which were dealt with in Section
14.2.
In this and the next two sections we'll discuss positive
series exclusively. They're series whose terms are positive or 0 (so their
associated sequences are positive sequences). As seen in Section
14.2 Theorem 3.2, a series converges iff its tail, no matter
where it begins, converges. Thus for the discussion to be more general we'll
discuss series that are ultimately positive. For the
definitions of (ultimately) positive and of (ultimately) increasing sequences,
see Section
14.1 Definitions 1.2 and Remarks 1.1
that follows it. Remember that the “ultimately” property includes the
corresponding normal property. For example, a positive
sequence (or series) is also an ultimately positive sequence (or series).
However an ultimately positive sequence (or series)
may or may not be a positive sequence (or series).
The proof of the comparison test that follows will make use
of a property of ultimately positive series, stated in the theorem
below. This theorem asserts that there are only two possibilities for an
ultimately positive series: it either converges or
diverges to infinity. If it diverges then it must diverge to infinity; it can't
diverge to negative infinity or simply diverge. Intuitively
this is “obvious”, as from some point on it can't decrease since all of its
terms from that point on are positive or 0; adding a
quantity that's positive or 0 to something can't decrease that something.
Theorem 1.1 – A
Property Of Ultimately Positive Series

Proof
EOP
^{{1.1}}
Section
14.1 Completeness Property.
^{{1.2}}
Section
14.2 Theorem 3.2.
^{{1.3}}
Section
14.2 Theorem 3.2.
This is a comparison termbyterm from some point on of the
two series, and therefore is called the standard comparison
test, abbreviated as “SCT”.
Theorem 1.2 – The
Standard Comparison Test (SCT)

Proof
EOP
^{{1.4}}
Section
14.1 Theorem 7.1.
This standard comparison test for series is similar to the
standard comparison test for improper integrals as presented in
Section
11.3 Theorem 3.1.
Example 1.1
Test the convergence of this series:
Note
Parts a, b, and c produce k = 11, 3,010, and 40 respectively. We choose the “nicest” of them, k = 40 (part c).
Solution
We have:
EOS
It would lead to nowhere if we compared 1/(10n + 3,000) to 1/n.
It would be 1/(10n + 3,000) < 1/n for all n. The SCT
doesn't apply in this case, because the inequality goes in the wrong direction
as the series of 1/n diverges to
infinity.
Example 1.2
Determine the convergence or divergence of the following
series. If it converges, determine the interval between two
consecutive integers that contains its sum.
Note
We know that n^{2} gets very
large very fast, thus that 1/n^{2} gets very
small very fast, as n gets larger
and larger. This leads us to
feel that the series of 1/n^{2} is likely to
converge. Consequently we'll compare it with the convergent telescoping series
of
1/(n – 1)n,
where n starts from 2 up.
Solution
EOS
This inequality is useless because it goes in the wrong direction.
We would be forced to change our feeling about the behavior
of the series.

Example 1.3
Establish the convergence or divergence of the following series.
If it converges, determine the interval between two
consecutive integers that contains its sum.
Solution
EOS
Example 1.4
Find out whether the following series converges or diverges.
If it converges, determine the interval between two consecutive
integers that contains its sum.
Solution
EOS
A dejavu: multiplying a series that diverges to infinity by
a positive constant, no matter how small, yields a series that still
diverges to infinity. As an intuitive mnemonic device, multiplying infinity by
a positive constant, no matter how small, still
produces infinity.
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2. The Limit Comparison Test (LCT) 
Consider the series:
the limit is still a finite number, and the same conclusion holds.
Instead of comparing two series term by term, we can just
compare their tails near infinity, as we know that a series
converges iff its tail converges. Comparison of the tails is carried out by
evaluating the limit at infinity of the quotient of the
corresponding terms of the two series. If the limit exists or is infinity, it
tells us of the behavior of one series if that of the other
is known.
The test of comparing the tails of the series by
investigating the limit of the quotient of the corresponding terms of the
series is
known as the limit comparison test, abbreviated as “LCT”.
Theorem 2.1 – The
Limit Comparison Test (LCT)

Proof
a. For sufficiently large n we have:
EOP
^{{2.1}} Section 14.2 Corollary 4.2.
The proof of the LCT makes use of the SCT. The LCT can therefore be considered as a corollary of the SCT.
Example 2.1
Test the convergence/divergence of the series:
Solution
EOS
If we didn't have the feeling that, near infinity, the general
term of the given series behaves like 1/n^{2} and if we
compared it to
1/n, we would get:
Example 2.2
Determine the behavior of the series:
Solution
EOS
Since for large n, (3^{n}
+ 1)/(5^{n} – 100) behaves like 3^{n}/5^{n}
= (3/5)^{n} of which the series converges, we have
the intuitive feeling
that the given series converges. So we compare them. We feel that using the SCT
would be complicated. Thus we use the
LCT, which, as it turns out, serves us well.
Example 2.3
In Example 1.1 we found out that the series:
diverges to infinity by using the SCT. Now show that it diverges to infinity by utilizing the LCT.
Solution
EOS
Clearly the LCT is simpler than the SCT in this case. We
don't have to look for a fraction 1/kn to compare
1/(10n + 3,000) to.
We simply utilize the simplest of them, 1/n.
Example 2.4
Does the following series converge or diverge?
Note
Of course we feel that this series converges as near
infinity it behaves like 3/n^{2}. So let's use
the LCT and 1/n^{2} to try to show
that it converges.
Solution
We have:
EOS
This example shows that if one comparison test fails, the
other may work. Of course if one comparison test doesn't work, we'll
try the other.
Example 2.5
Establish the convergence or divergence of this series:
Solution
EOS
Before trying to think of a series to employ in a comparison
test, be alert for the possibility that the associated sequence of the
given series doesn't approach 0.
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3. Remarks 
Correct Feeling
About Convergence Or Divergence
All of the examples above give us the idea that to test the
convergence of a series successfully without wasting time, and time
is precious in a test or an exam, we should have a correct feeling about its
convergence or divergence. If we feel that a series
is likely to converge, we'll use a convergent series for comparison, and if we
feel that it's likely to diverge, we'll use a divergent
series. If, for a given series, one comparison test is complicated or fails,
we'll try the other, or one of the other ones presented
in later sections.
Don't forget to mentally examine the limit of the terms of
the series to see whether it's 0 or not before attempting to apply a
convergence test. If it doesn't seem to be 0, evaluate it. Of course if it's
not 0 then the series must diverge.
Things Involved
In The Comparison Tests
The two comparison tests involve two series, one whose
behavior is to be determined and the other whose behavior is known.
The tests compare the terms of the former to those of the latter.
Problems & Solutions 
1. For each of the following statements, mark it as T (True) or F (False).
Solution
2. Determine the convergence or divergence of the following series. Find its sum if it converges.
Solution
Note
3. Test the convergence of the following series.
Solution
The two series on the righthand side are geometric series
with common ratios 5/8 and 7/8 respectively, and –1 < 5/8 < 1
and
–1 < 7/8 < 1. It follows that they both converge. Therefore the given
series converges.
Note
Part b reminds us that before trying to test a series, we
should examine mentally the limit of its terms. If the limit doesn't
seem to be 0, we should calculate it. Parts b and c don't utilize any test at
all.
4. Determine the convergence/divergence of these series:
Solution
5. Establish the convergence or divergence of the following series:
Solution
6. Does the following series converge or diverge?
Solution
7. Find out whether the following series converges or diverges:
Solution
8. Show that this series:
Solution
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