In this chapter we will study an important convergence criterion called the direct comparison test. This criterion allows us to deduce the convergence behavior of a series by comparing it to that of another series. That way we can answer questions regarding the convergence of a series by considering a simpler series. Using this criterion we can estimate upper or lower bounds of the series, until we hopefully find a proof for the convergence behavior.
The direct comparison test is also used to proof other criteria such as the quotient test and the root test, which are both useful tools to have when solving problems about the convergence of series.
Direct comparison test: majorant[Bearbeiten]
There are two versions of the direct comparison test. The first one is a test for convergence and involves finding a majorant of the series:
Theorem (Direct comparison: majorant)
Let be a series. If there exists a convergent series with for all then converges absolutely.
Note that from the inequality it automatically follows that , because is greater or equal to the non negative number .
Proof (Direct comparison: majorant)
If converges, then the sequences of partial sums is bounded above . Because for all also the corresponding sequence of partial sums of has an upper bound:
The sequence of partial sums is monotonically increasing since . We have that the series is monotonic and bounded. Therefore is convergent.
Comprehension question: Would it suffice for the majorant if there exists so that for all ?
Yes. If the sequence of partial sums is bounded above, then this is true also for . Because of
also the sequence of partial sums is bounded above. Since the finitely many summands don't change the boundedness, the partial sums of also have an upper bound. This means that converges absolutely.
As we already mentioned in the introduction, we would like to find a preferably simple convergent series to use as our majorant. Often, we can use the series as our majorant. We could also use a more general harmonic series for . Another good option is the convergent geometric series for , for example .
Direct comparison test: minorant[Bearbeiten]
The second comparison test is similar, but we use it to determine the divergence of a series, and it involves finding a minorant of that series.
Theorem (Direct comparison test: minorant)
Let be a series with for all . If there exists a divergent series with for all , then also the series is divergent.
Proof (Direct comparison test: minorant)
Since the sequence of partial sums of is monotonic. According to our premise this series is divergent, so it must be unbounded. Because for all we have that for all and therefore also the sequence of partial sums of is unbounded. This implies that diverges (every unbounded sequence diverges.)
Analogue to the majorant condition it sufficies that be true for all , for a fixed .
When using the minorant criterion in practice, often a good choice for the minorant is the divergent harmonic series . But also the series for (for example ) make for good minorants. Also the geometric series with is suited as a minorant.
When applying the direct comparison test using a minorant it is necessary that ! If we only have that for as in the majorant case, then we cannot follow the divergence of from the divergence of . We can only say that doesn't converge absolutely. To see this consider for example the series and with and . We have , and the harmonic series diverges. But we can show that converges according to the Leibniz-Kriterium.
Examples and exercises[Bearbeiten]
Direct comparison with majorant[Bearbeiten]
Example (Direct comparison with majorant)
Take the series . The corresponding sequence is . If we factor out in the denominator and reduce we get:
The series behaves like and should therefore converge. We can prove this formally using our direct comparison test: From we have that
With we have found a majorant that converges. From our direct comparison test it follows that also converges.
Exercise (Direct comparison with majorant)
Examine whether the series converges.
How to get to the proof? (Direct comparison with majorant)
If one has little experience proving convergence of series, it is difficult to tell whether the series is convergent or divergent. Start by examining the fraction . Maybe you see that the numerator is a polynomial of degree two and the denominator is a polynomial of degree three. This means that the whole fraction goes to zero with a rate of convergence . In our series this fraction is squared. This means that will go to zero with a convergences rate of . Since is convergent, also the series should converge.
To prove this formally, we can use the direct comparison test and compare to the majorant . But we first need to show that this is indeed a majorant of our series. Here we need to make some clever estimates:
To estimate an upper bound we followed a schema: We removed summands that make the term smaller. For the remaining summands, that we could not simply remove, we found an estimate so that we could merge them. The end result:
Thus is indeed a majorant, and since the series is convergent, by direct comparison we find that also must be convergent.
Solution (Direct comparison with majorant)
We have that:
is a convergent majorant of the series . By direct comparison the series is convergent.
Direct comparison with minorant[Bearbeiten]
Example (Direct comparison with minorant)
Let us examine the series . It should grow similar to . Since the harmonic series diverges, also the series should diverges, and thus also our original series. Formal proof:
Thus is a minorant and diverges. From direct comparison it follows that is also divergent.
Exercise (Direct comparison with minorant)
Examine whether the series converges or diverges.
How to get to the proof? (Direct comparison with minorant)
Here it is interesting to note the rate at which the summands converge to zero. The product converges like towards infinity. Thus converges to 0 like . Since we take the square root we can see that the rate of convergence of is like . But the harmonic series is divergent. This tells us that should also diverge.
Now that we have conjectured the divergence of the series we need to prove it. We do that by direct comparison with a minorant. First we need to estimate lower bounds:
As in the previous exercise when we needed to estimate an upper bound, we can follow a schema: Summands that make the term bigger, can be removed. For the remaining summands we try to find an estimate so that we can merge them. So now we have a divergent minorant: . By direct comparison it follows that is divergent.
Proof (Direct comparison with minorant)
The series is divergent. Thus also must diverge by direct comparison.
Corollary: Limit comparison test[Bearbeiten]
For series with positive summands there is another test called the limit comparison test, which we can derive form the direct comparison test:
Theorem (Limit comparison test)
Let and be series with positive summands, and with , then the series converges if and only if converges.
Proof (Limit comparison test)
Because there exists , so that for all . This follows from the epsilon definition of convergence with . For all we have:
If converges, then also converges by direct comparison because of the inequality for all . We also see that . But then also converges, because multiplying by a constant factor doesn't affect convergence behaviour.
On the other side assume that is convergent, then also is convergent (again multiplication by a constant factor). Because of the inequality for all the series is convergent by direct comparison.
Comprehension question: What can we deduce from under the same conditions?
From the convergence of we can deduce the convergence of . If , there exists with for all . Thus if is convergent it follows from direct comparison that is convergent.
The opposite is not true: If converges we cannot conclude that is also convergent. As a counterexample consider and . Here we have
But is convergent and is divergent .
Example (Prove convergence using limit comparison test)
Consider the series . We have
With it follows that
Since is convergent, is follows from the limit comparison test that is also convergent .
As this example shows, the limit comparison test is a convenient way to prove the convergence of a series. Often, computing the limit is easier than finding a suiting majorant. In many lectures however the limit comparison test is not discussed, so you are probably not allowed to use it either. Even so you can still use the argumentation of the proof to find the necessary majorant:
For and we find that . There must exist with for all . This is equivalent to for all . This way we have found a suitable majorant. Because we know that converges, also must converge. Now we use the direct comparison test to show that is convergent.
Exercise (Limit comparison test)
Examine for which the series converges.
Hint: Consider the general harmonic series .
Solution (Limit comparison test)
Here is not a null sequence:
Let and . We have
Because of and from the Grenzwertsätze it follows . According to the limit comparison test, the series converges if and only if converges. Thus we know it converges for and diverges for .
Combining both cases we find that our series is convergent for and divergent for .