In this section we will be comparing a given series with series that we know either converge or diverge.
Let and be positive sequences where for all , for some .
If converges, then converges.
If diverges, then diverges.
First consider the partial sums of each series.
Since both series have positive terms we know that
and
Therefore, both of the sequences of partial sums, and , are increasing.
For , We’re now going to split each series into two parts:
This means that and . Also, because for all , we must have .
For the first part of the theorem, assume that converges. Since we know that
From above we know that for all so we also have
Because converges it must have a finite value and is bounded above. We also showed that is increasing so by Theorem 9.1.6 we know converges and so converges.
For the second part, assume that diverges. Because we must have . We also know that for all , . This means that
Therefore, is a divergent sequence and so diverges. ∎
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Direct Comparison Test / Limit Comparison Test for Series — Basic Info from https://youtu.be/LAHKu3B-1zE
Determine the convergence of .
SolutionThis series is neither a geometric or -series, but seems related. We predict it will converge, so we look for a series with larger terms that converges. (Note too that the Integral Test seems difficult to apply here.)
Since , for all . The series is a convergent geometric series; by Theorem 9.4.1, converges.
Determine the convergence of .
SolutionWe know the Harmonic Series diverges, and it seems that the given series is closely related to it, hence we predict it will diverge.
We have for all .
The Harmonic Series, , diverges, so we conclude that diverges as well.
The concept of direct comparison is powerful and often relatively easy to apply. Practice helps one develop the necessary intuition to quickly pick a proper series with which to compare. However, it is easy to construct a series for which it is difficult to apply the Direct Comparison Test.
Consider . It is very similar to the divergent series given in Example 9.4.2. We suspect that it also diverges, as for large . However, the inequality that we naturally want to use “goes the wrong way”: since for all . The given series has terms less than the terms of a divergent series, and we cannot conclude anything from this.
Fortunately, we can apply another test to the given series to determine its convergence.
Let and be positive sequences.
If , where is a positive real number, then and either both converge or both diverge.
If , then if converges, then so does .
If , then if diverges, then so does .
We have so we can find two positive numbers, and such that . Because we know that for large enough the quotient must be close to . So there must be a positive integer such that if we also have . Multiply by and we have for . If diverges, then so does . Also since for sufficiently large , by the Comparison Test also diverges.
Similarly, if converges, then so does . Since for sufficiently large , by the Comparison Test also converges.
Since , there is a number such that
Now since converges, converges by the Comparison Test.
Since , there is a number such that
Now since diverges, diverges by the Comparison Test.∎
Theorem 9.4.2 is most useful when the convergence of the series from is known and we are trying to determine the convergence of the series from .
We use the Limit Comparison Test in the next example to examine the series which motivated this new test.
Determine the convergence of using the Limit Comparison Test.
SolutionWe compare the terms of to the terms of the Harmonic Sequence :
Since the Harmonic Series diverges, we conclude that diverges as well.
Determine the convergence of
SolutionThis series is similar to the one in Example 9.4.1, but now we are considering “” instead of “.” This difference makes applying the Direct Comparison Test difficult.
Instead, we use the Limit Comparison Test with the series :
We know is a convergent geometric series, hence converges as well.
As mentioned before, practice helps one develop the intuition to quickly choose a series with which to compare. A general rule of thumb is to pick a series based on the dominant term in the expression of . It is also helpful to note that factorials dominate increasing exponentials, which dominate algebraic functions (e.g., polynomials), which dominate logarithms. In the previous example, the dominant term of was , so we compared the series to . It is hard to apply the Limit Comparison Test to series containing factorials, though, as we have not learned how to apply L’Hôpital’s Rule to .
Determine the convergence of .
SolutionWe naïvely attempt to apply the rule of thumb given above and note that the dominant term in the expression of the series is . Knowing that converges, we attempt to apply the Limit Comparison Test:
Theorem 9.4.2 part (3) only applies when diverges; in our case, it converges. Ultimately, our test has not revealed anything about the convergence of our series.
The problem is that we chose a poor series with which to compare. Since the numerator and denominator of the terms of the series are both algebraic functions, we should have compared our series to the dominant term of the numerator divided by the dominant term of the denominator.
The dominant term of the numerator is and the dominant term of the denominator is . Thus we should compare the terms of the given series to :
Since the -series converges, we conclude that converges as well.
The tests we have encountered so far have required that we analyze series from positive sequences (the absolute value of the ratio and the root tests of Section 9.6 will convert the sequence into a positive sequence). The next section relaxes this restriction by considering alternating series, where the underlying sequence has terms that alternate between being positive and negative.
Suppose is convergent, and there are sequences and such that for all . What can be said about the series and ?
Suppose is divergent, and there are sequences and such that for all . What can be said about the series and ?
In Exercises 3–8., use the Direct Comparison Test to determine the convergence of the given series; state what series is used for comparison.
In Exercises 9–14., use the Limit Comparison Test to determine the convergence of the given series; state what series is used for comparison.
In Exercises 15–24., use the Direct Comparison Test or the Limit Comparison Test to determine the convergence of the given series. State which series is used for comparison.
In Exercises 25–32., determine the convergence of the given series. State the test used; more than one test may be appropriate.
We have shown that the harmonic series diverges. Suppose we remove some terms by considering the series where is the prime (so , , , etc). Determine if this series converges or diverges, using the fact that for all sufficiently large.