Exercises: Convergence criteria for series – Serlo

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Application of convergence criteria[Bearbeiten]

Exercise (Convergence proof training 1)

Investigate, whether the following series are divergent, convergent or even absolutely convergent.

Solution (Convergence proof training 1)

1. Root test: The in the exponents is tedious. we can remove it from one exponent by taking the -th root:

The root test now tells us that the series converges absolutely. Intuitively, the exponent produces a much faster growth than . So fast that decays faster than a geometric series .

2. Ratio test: By which factor do the sequence elements increase, if we go from to ? If that factor is lower than some constant , we have absolute convergence:

So the increase factor is smaller than and the series converges absolutely.

3. Diract comparison test: For large , the series elements essentially behave like , which is a divergent harmonic series. So we suspect that this series also diverges. In fact, the in the denominator makes smaller than . But for , will approach it arbitrarily close and get larger that for any . For convenience, we choose and get that eventually:

  • But the harmonic series diverges.

The series is "even larger" and therefore also diverges.

4. Term test: What happens for ? Actually, the series elements get close to a constant:

The series would intuitively evaluate to and is in mathematical words divergent by the term test.

5. Root test: For large , we have seen in 4., that goes like . So should behave like , which is a geometric series and hence convergent. We can verify this convergence by the root criterion:

Hence, the series converges absolutely.

6. Alternating series test:

The suggests that we have to deal with an alternating series. And since is always positive, this is indeed true. Further, the two square roots are moving very close together for high . For instance , but and (you may verify this yourself by taking the first-order Taylor approximation or a pocket calculator). We therefore suspect to be monotonously decreasing, which we will prove in the following:

So

  • is indeed monotonously decreasing, since
  • is a null sequence, since .

So the series converges.

However, the series does not converge absolutely, since we can write as a divergent telescoping series

7. Term test: For large , the expression converges to 1. So we expect the absolute value of to converge to 1 and the series should diverge by means of the term test. Now, every second element is negative, so we have to restrict to a subsequence of positive elements:

Indeed, the subsequence of doesn't converge to 0, so cannot be a null sequence and by means of the term test, the series is divergent.

Note: The may tempt one to assume that the alternating series test is a good tool to use. However, the alternating series test can only prove convergence and as the series is divergent, any attempt using it bound to fail! Indeed, is not a null sequence, which is why the alternating series test doesn't work.

8. alternating series test: The converges to , so the series elements behave like for large . This is an alternating harmonic series, so we suspect to converge, but not absolutely. We prove this using the alternating series test: For there is

  • , so is monotonously decreasing.
  • , since . So is a null sequence.

Therefore, the series converges.

To show that it does not converge absolutely, we compare to a smaller but still divergent harmonic series (direct comparison test):

  • , since is monotonously increasing.
  • still diverges. (harmonic series)

And hence, the series of absolute values diverges.

Exercise (Convergence proof training 2)

Investigate, whether the following series are divergent, convergent or even absolutely convergent.

Solution (Convergence proof training 2)

1. Direct comparison test: It is useful to know that eventually grows slower than any of the polynomials , i.e. any with . We can generously take and estimate  :

So the series elements grow slower than and the series converges absolutely.

Note: For , there is and we could even get .


2. Direct comparison test: Since , the series elements grow slower than

  • but diverges (harmonic series)

So the series diverges.

3. Ratio test: We write . Both and grow fast at high . But eventually, the exponential "wins" against all polynomial functions ,such that and even . Mathematically, this can be proven by considering the ratio between two subsequent elements:

So the series converges absolutely.

Note: We could have done the same proof with for any (not just ). The exponential always "wins" against the polynomial.

Alternative: Root est:

So the series converges absolutely.

4. Term test: For , the approaches 1 as . We select a subsequence consisting of only the positive entries:

Therefore, cannot be a null sequence and the corresponding series diverges.

Note: The may trick you into taking the alternating series test for proving that the series converges. However, this doesn't work since is not a null sequence!

5. Alternating series test: This time, , as . So we have a in front of a null sequence and may try the alternating series test:

  • , since is monotonously decreasing. Therefore, is also monotonously decreasing.
  • , since is continuous. Therefore, is a null sequence.

By the alternating series test, the series converges.

Note: The series does not converge absolutely, since for large , there is . And is a divergent harmonic series.

6. Direct comparison test: The can be written as . So it decays like (a geometric series). We compare to this geometric series

  • , since .
  • (geometric series)

Therefore, the series converges absolutely.

7. Direct comparison test: The is oscillating with amplitude (therefore staying bounded) and the decays fast enough to get convergence. We can hence compare:

So the series converges absolutely.

Note: Even though the series is oscillating, the alternating series test doesn't work here, since is not monotonously decreasing!

Exercise (Series depending on parameters)

The following series depend on a parameter . For some , they will converge, for others not. Your job is to investigate for which , the series converges (absolutely) or diverges:

Solution (Series depending on parameters)

1. Ratio test: Intuitively, a factorial grows faster than any exponential in a way that not just , but even the series converges. This should also hold true if we replace by the larger expression . We can verify this mathematically by the ratio test:

So series is absolutely convergent for all .

2. Case distinction:

Fall 1:

The series elements are smaller than , which is known to be a convergent series:

So by the direct comparison test, the series converges absolutely.

Fall 2:

In this case, we expect the to "explode exponentially" in , while is growing only polynomially (and hence slower). So the series should converge. Mathematically, this can be verified by the root test: , da

So the series diverges.

3. Case distinction:

Wir unterscheiden zwei Fälle:

Fall 1:

In that case, the decays exponentially, and therefore "wins" against the polynomially increasing . We use the ratio test to verify this:

So indeed, the series diverges absolutely.

Fall 2:

Here, the is constant () or exponentially increasing. Multiplying it by we get sequence elements which are increasing and hence bigger than a constant. This suggests using the term test: . So is not a null sequence.

Therefore, the series diverges.

4. Case distinction:

Fall 1:

Here, and (geometric series)

So by the direct comparison test, the series converges absolutely.

Fall 2:

, which is a divergent harmonic series.

Fall 3:

is an alternating harmonic series, which converges but not absolutely, by the alternating series test.

Fall 4:

We have , where the geometric series diverges (direct comparison test).

Note: The cases and can also be treated by the root test or the ratio test.

Exercise (Direct comparison in asymptotic behaviour)

Investigate, whether the series

converges or diverges.

Solution (Direct comparison in asymptotic behaviour)

We take a look at the asymptotic behaviour for :

So it might be useful to compare with a series scaling like . In fact,

This convergence tells us that for high , there must be a with

for all

So we can compare to . Since the series converges, we know by the direct comparison test that also converges absolutely.

Term test - strengthened version[Bearbeiten]

Exercise (Term test - strengthened version)

The term test tells us that if is monotonously decreasing and converges, then is a null sequence. Prove that in this case, even must be a null sequence. (This means that it suffices to show that is not a null sequence in order to conclude that diverges.)

Solution (Term test - strengthened version)

The intuition behind this test is that being not a null sequence means that decays slower than or as fast as a harmonic sequence , and diverges. A formal proof can be done by using the Cauchy criterion: Assume that converges. then the sequence of partial sums is a Cauchy sequence. We will first use this fact to bound all even elements of :

Proof step: is a null sequence

Since is a Cauchy sequence, there must be an and an , such that for all there is

Hence, for all :

So we know that and hence also eventually falls below each , so they are null sequences.

Now we are still missing the odd numbers:

Proof step: is a null sequence

The argumentation is the same as above: is a Cauchy sequence, so for each there is an , such that for all there is

Hence, for all :

So we know that and hence also are null sequences.

The case distinction was actually only necessary since for even numbers , we had to sum over and for odd numbers over elements. Since both for even and odd indices , the elements tend to zero as , the sequence must also tend to zero, i.e. it is a null sequence.

Cauchy criterion: an application[Bearbeiten]

Exercise (Alternating harmonic series)

Using the Cauchy criterion, prove that the alternating harmonic series converges. (note: the harmonic series does not converge!)

Solution (Alternating harmonic series)

Our aim is to show that the sequence of partial sums is a Cauchy sequence. That means, we have to put a bound on

The sequence is definitely null. So for each we can choose an , such that for all . So is a Cauchy sequence implying that converges, which was to be shown.

Root and ratio test: estimate of errors[Bearbeiten]

In some cases, the infinite sum cannot be computed explicitly. We can try to approximate it by . The following exercises are devoted to giving upper bounds on the approximation error :

Exercise (Error estimate for the root test)

Let be a sequence and . Further, for all . Then, the series converges absolutely by the root test. Show that it can be approximated by with up to a maximal error of

Solution (Error estimate for the root test)

The idea is that the root test bounds by a geometric series and This bounding can be explicitly shown by using the root test criterion:

So for , we can sum up:

Exercise (Error estimate for the ratio test)

Let be a sequence and . Further, let and for all . By the ratio test, the series converges absolutely. Show that it can be approximated by with up to a maximal error of

Solution (Error estimate for the ratio test)

Again, we bound the series by a geometric series. But this time, the factor of is replaced by . By assumption:

For any , we can conclude by iteration:

Again, we sum up the geometric series bounds:

Exercise (showing that a sequence is null)

  1. Let be a sequence and . Further, let and or . Show that we then have a null sequence .
  2. Conclude that for and .

Solution (showing that a sequence is null)

  1. Both (ratio test) and (root test) imply that the series converges. But by means of the term test , the sequence has to be a null sequence.
  2. We specifically consider the sequence . It satisfies the ratio test criterion:

So the series converges and by part 1 of the exercise, is a null sequence, i.e. .

Alternating series test with error bounds[Bearbeiten]

Exercise (Alternating series test with error bounds)

First, show that the series

converges. Then, determine an index , such that the partial sum for approximates up to a precision of \tfrac{1}{100}</math> .

Solution (Alternating series test with error bounds)

Proof step: The series converges

For we can estimate

So is monotonously decreasing. Further

So is even a null sequence. By means of the alternating series test, the series hence converges.

Proof step: Finding

For the alternating series test, there is an error bound:

With . It saves a ot of calculation work to just give a coarse upper bound for this expression:

Now, if , then there is also . Hence,

So (and any bigger ) leads to a precise enough approximation. For the approximation precision is therefore better than .

Cauchy condensation test[Bearbeiten]

Exercise (A double logarithm)

Determine, for which the following series converges:

Solution (A double logarithm)

The Cauchy condensation test is useful to resolve the double logarithm: if in we only consider the elements (i.e. 2, 4, 8, 16, ...), we get obtain a single logarithm in instead of a double logarithm: . As in our sequence, is monotonously decreasing, the Cauchy condensation test yields that it converges if and only if the following series converges:

In the article "Cauchy condensation test", there is an exercise proving that converges for and diverges for . This can be used for a direct comparison:

Fall 1:

Here,

and . So by direct comparison, our series converges for all .

Fall 2:

Here,

and diverges. So by direct comparison, the series diverges for all .

Further convergence criteria[Bearbeiten]

Exercise (Series with products)

Let and be two real sequences. Prove:

  1. Whenever the series converges absolutely and is bounded, then also converges absolutely.
  2. There are pairs of a series , which converges and a sequence which is bounded, such that does not converge. So the above statement does only hold true for absolute convergence, but not for convergence in general.

Solution (Series with products)

Part 1:

Way 1: Boundedness of partial sums

Since converges absolutely, the sequence of partial sums must be bounded. Further, is bounded. That means, there is an upper bound with for all and therefore

But now, is bounded, so is bounded as well. The sequence is smaller than and therefore also bounded, i.e. converges absolutely.

Way 2: Direct comparison test

As above, has an upper bound , meaning that for all . Hence,

Now, the series converges, as well as . We use the direct comparison test: is smaller than and hence converges.

Part 2: We know that the harmonic series diverges, but can be made convergent by putting a minus sign in front of every second element . The idea is now to "reverse" this process: we take the convergent alternating series and make it again divergent turning every second element positive:

The "reversing sequence" is bounded, as . So converges and is bounded, but diverges as a harmonic series.

Note: other counter-examples are also possible.

Exercise (Raabe criterion)

  1. Let and be two real sequences. Prove: If for almost all , there is (all except for a finite number of elements). Then, if
    • for some , then converges absolutely.
    • then diverges.
  2. Use this so-called Raabe criterion, to prove that the following series for any  :

Solution (Raabe criterion)

Part 1:

  • For convergence, we equivalently transform

This works for almost all . So we can find an , such that the re-formulation works for all . We take the sum of both sides starting from up to some :

Therefore, the sequence of partial sums is bounded. That means, the series converges absolutely. Adding a finite number of elements, we get that also converges absolutely.

  • In the case, where divergence should hold, we can equivalently transform for all  :

This is equivalent to

The harmonic series diverges. By direct comparison, this implies that also diverges. Adding a finite number of elements, also diverges, which was to be shown.

Part 2: Here, , which means

So is a suitable constant to use the Raabe criterion and imply absolute convergence of the series .