Exercises: convergence and divergence – Serlo

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Proving convergence of a sequence[Bearbeiten]

Exercise (Convergence of a sequence 1)

Use the epsilon-definition to prove that the sequence converges. What is its limit?

How to get to the proof? (Convergence of a sequence 1)

How does this sequence behave for large ? The denominator stays constant at , while the enumerator grows towards infinity. So the fraction should go to .

Now, let us prove this. The definition of the limit for this sequence reads:

Now,

The Archimedean Axiom allows to find an with . By the above inequalities, for all sequence elements with , there is . So the sequence converges to 0.

Solution (Convergence of a sequence 1)

Let . We choose an with . This is possible by the Archimedean axiom. Then, for all there is

Exercise (Convergence of a sequence 2)

Use the epsilon-definition to show that the sequence converges to .

How to get to the proof? (Convergence of a sequence 2)

For the epsilon-definition, we need to verify:

This means, for each , we need to find a suitable , such that for holds after the sequence has passed element number . So let us find such an by re-shaping the condition:

That means, we need an with . This exists by the Archimedean axiom and we are done with the proof.

Solution (Convergence of a sequence 2)

Let . We choose an with . This can be done by the Archimedean axiom. Then, for all there is

Proving divergence for an alternating sequence [Bearbeiten]

Exercise (Divergence of an alternating sequence)

Prove that the sequence diverges.

How to get to the proof? (Divergence of an alternating sequence)

We need to show that

Intuitively, as the sequence alternates between and , the only possible limits are and . However, we have to show the above statement for all in order to prove divergence.

First, let us consider the case : For all even there is , so

That means, we are further away from than . For any there will be an even with , such that

Analogously, for and all odd :

, since

So if we choose again , then for each there will be an odd with , such that

I.e. for all , the sequence has infinitely many elements staying away by more than 1 from and hence, the sequence diverges.

Proof (Divergence of an alternating sequence)

Case 1: Let . Choose and arbitrary. Now, take an even with . Then,

Case 2: Let . Choose and arbitrary. Now, take an odd with . Then,

Proof (Alternative proof by contradiction)

For two subsequence elements and there is always . So if for there was an and an with

then, we would have

Powers of sequences[Bearbeiten]

Exercise (Powers of sequences)

For any power and a given sequence converging as the -th power of this sequence will also converge to . Prove this statement directly, using the epsilon-definition (without the product rule).

How to get to the proof? (Powers of sequences)

We need to show that

will fall below any . What we can use is that gets arbitrarily small. It is therefore useful, to get a factor of out of the term . This is done by using

We can prove this statement similar to the geometric sum formula:

We use this auxiliary formula with and :

The first factor of is nice, since we can control it. The second factor will be bounded by the triangle inequality and we get

If we can now show that the second factor is bounded from above by a constant, then gets arbitrarily small and we are done. Now, we know that converges, so it is bounded. We call this bound with for all . Therefore, .

Therefore, gets arbitrarily small (below any ) and we have convergence .

Proof (Powers of sequences)

Let be arbitrary. Since converges, there must be an with for all . Since there is an with for all , where we define the constant . Let now be arbitrary. Then

Limit theorems[Bearbeiten]

Exercise (Limits of sequences)

Investigate, whether , , , , and converge. Determine the limit in the respective case.

Solution (Limits of sequences)

1. The trick is to split the roots and in factors and . We can then use that and and patch them together, using the limit theorems:

2. For fractions of polynomials, the "polynomial with highest degree" usually wins. We have two polynomials of equal degree 3. In that case, one has a finite limit. we can determine it by factoring out the highest power and using the limit theorems.

3. It looks as if the enumerator was of degree 2 and the denominator of degree 1, so one might think at first glance that the enumerator "wins" and we get divergence to infinity. However, there is a minus sign in the enumerator, so we need to do some simplification first. indeed, the -terms cancel and we get a finite limit:

4. What will happen for large ? At first glance, we go to which is no determinate result. but after some simplifications:

Alternative solution: (squeeze theorem)

There is . The squeeze theorem hence yields .

5. Again a square root. For large , there will be So both enumerator and denominator behave like a polynomial of degree 1. We factor out an and obtain a finite limit:

6. The long sum can be simplified using Gauss' sum formula . Then, we get two polynomials of degree 2, where we can factor out an :

Warning

It is tempting to think: " is a polynomial of degree 1" and to factor out an . This will not yield the desired result, as the number of summands in this polynomial depends in . If you have a sum whose number of elements increases in , you must always resolve the sum first into some expression with a finite amount of terms! There are cases in which you have sums, where each summand goes to 0, but you get Fehler beim Parsen (Unbekannte Funktion „\intfy“): {\displaystyle \intfy } -many summands, so the limit is not necessarily 0.

Exercise (Sequences depending on a parameter)

Investigate whether with converges, depending on the parameter .

How to get to the proof? (Sequences depending on a parameter)

The result will depend on whether or . In case , will stay bounded and dominates both the enumerator and the denominator. In case , grows exponentially and hence dominates the fraction.

Proof (Sequences depending on a parameter)

Case 1: For there is

,

since .

Case 2: For there is

,

since .

Exponential sequences[Bearbeiten]

Exercise (Exponential sequences)

Investigate the following sequences , , and for converges. If they converge, determine the limit.

Solution (Exponential sequences)

1.We make use of :

2. As above, by :

3. This case requires a little index shift

4. In order to apply , we will extend both the enumerator and the denominator by

,

da

Alternative solution: If we already know that , then we directly get

,

Squeeze theorem[Bearbeiten]

Exercise (Squeeze theorem)

Determine the limit of the following sequences :

  1. for
  2. for
  3. for

Solution (Squeeze theorem)

1. For we have the upper and lower bound

both tend to . The squeeze theorem now implies .

2. For there is , which gives an upper bound. For the lower bound, we just use 0:

Since the squeeze theorem renders .

3. For we can find some with , i.e. we bound by the next smaller or bigger natural number. Since the root is monotonous, we have as an upper or lower bound:

The limit theorems now yield and . So the original sequence is "squeezed" towards .

4. For , there is

The limits of the bounding sequences are and . So by the squeeze theorem .

In case we analogously obtain .

Both cases can be concluded as .

5. The powers 2 in the enumerator and denominator become relatively small compared to the powers 3 and should hence not affect the convergence. We establish an upper and a lower bounding sequence by leaving them out

Those bounding sequences converge to

and

So by means of the squeeze theorem .

6. There is

Both bounds converge to

So by the squeeze theorem . This trick would also work, if we replaced 4 by any other natural number .

Exercise (Squeeze theorem for products of roots)

Prove that .

Solution (Squeeze theorem for products of roots)

It is easy to see that the sequence elements are greater than 0. So we need an upper bounding sequence for which converges to 0. That means, we have to show that goes to 0 fast enough. More explicitly, we need that decreases as a sufficiently large power of . We consider :

This implies

We can replace the by a sufficient power of

So there is also an upper bounding sequence converging to 0 and by the squeeze theorem .

Monotony criterion and recursively defined sequences[Bearbeiten]

Exercise (Monotony criterion)

Let be a sequence with for all . Show, using the monotony criterion that in this case the product sequence defined as

converges.

Solution (Monotony criterion)

This can be seen as a recursively defined sequence with , but one where we know the explicit form.

Step 1: is monotonously decreasing, i.e. .

This is easy to see, as sequence elements tend to get smaller:

Step 2: is bounded from below by , i.e .

This can be shown by induction in :

Induction basis: . .

Induction step:

Step 3: actually converges.

This is a direct consequence of the monotony criterion: is monotonously decreasing and bounded, so it converges.

Exercise (Convergence of a recursively defined sequence 1)

Why does the recursively defined sequence

converge? Determine its limit:

  1. By finding an explicit form of the sequence
  2. Using the monotony criterion

Solution (Convergence of a recursively defined sequence 1)

Part 1: Let us investigate the first sequence elements

Can we find a pattern? With a bit of training, one might see:

So we assert for all . Let us prove the assertion by induction over :

Induction basis: . .

Induction step:

This verifies our assertion for all (even though it was not easy to find). We can directly compute the limit of this expression using the limit theorems:

Part 2: is solved in 4 steps:

Step 1: is monotonously increasing, i.e. .

The proof runs by induction over :

Induction basis: . .

Induction step:

Step 2: is bounded from above by , i.e. .

This can also be shown inductively:

Induction basis: . .

Induction step:

Step 3: converges.

As is increasing and bounden, the monotony criterion can be applied and converges.

Step 4: computing the limit.

An index shift will not affect the limit: . But plugging in the recursion formula into this equation and using the limit theorems , we will obtain the limit:

We resolve for :

And the limit is (same as in part 1).

Exercise (convergence of a recursively defined sequence 2)

Let . Why does the recursively defined sequence

converge? What is its limit?

Solution (convergence of a recursively defined sequence 2)

There is

We apply this step times and get

So all differences are multiples of We obtain by adding up all differences (telescoping sum)

and get a geometric series:

There is so the limit theorems yield

Exercise (Monotony criterion for sequences)

Show the the sequence defined by the recursion relation

converges to the so-called golden ratio .

How to get to the proof? (Monotony criterion for sequences)

At first, we need a proof of convergence. Then, we can compute the limit. So first, let us show that is monotonous and bounded (then, the monotony criterion can be applied).

Why is it monotonous? Let us compute the first sequence elements:

We assert that is monotonously increasing. Only a proof is required, e.g. by induction. Then, we need an upper bound. The limit of shall be , so any number greater than will be an upper bound. Since w simply choose as an upper bound for . Of course, we need to verify that this is an upper bound by induction.

At this point, we have convergence of the sequence. Then, we need to show that is the limit and we are done.

Solution (Monotony criterion for sequences)

Step 1: is monotonously increasing, i.e. .

The proof is by induction over :

Induction basis: . .

Induction step:

Step 2: is bounded from above by , i.e. .

The proof is again done by induction:

Induction basis: . .

Induction step:

Note: We can also directly show by induction - if we first establish the equality .

Step 3: converges.

is monotonously increasing and bounded, so it converges by the monotony criterion.

Step 4: computing the limit.

An index shift will not change the limit: . We use the recursive definition, as well as the limit theorems and continuity of the root function

Now, we resolve for and obtain

Which has two solutions:

I.e. and . Which one is the limit? There is . Since and monotony, the limit of can only be above . Hence, can not be the limit and we have convergence to .

So we have established convergence to .

Exercise (Monotony criterion for sequences)

Show that the sequence , defined by

converges to the golden ratio . Do this by showing

  1. for all
  2. The subsequence is monotonously increasing and the subsequence is monotonously decreasing.
  3. Both subsequences converge to the same limit .

Solution (Monotony criterion for sequences)

Part 1: is done by induction over :

Induction basis: .

Induction step:

Part 2: is also done by induction over :

  • At first, we show that the odd elements are monotonously increasing, i.e. for all .

Induction basis: .

Induction step:

  • Now, monotonous decrease for even elements is shown, meaning for all .

Induction basis: .

Induction step:

Part 3: By means of parts 1 and 2, we have that is monotonously decreasing, is monotonously increasing and both are bounded. The monotony criterion implies convergence of both subsequences.

Now, , since , so there is

We resolve the quadratic equation, obtaining two solutions:

and

By means of part 1, there is , and hence . So the latter solution must be the limit: .

Analogously, . Both subsequences converge to the same limit, so there must be

which finishes the proof.

Cauchy's limit theorem and the Cesàro mean[Bearbeiten]

Exercise (Cauchy's limit theorem)

  1. Prove Cauchy's limit theorem: Let be a sequence converging to . Then, the Cesàro mean converges to .
  2. Does the converse also hold true? I.e. if there is does this imply  ?
  3. Use step 1 to show: .

Solution (Cauchy's limit theorem)

  1. Since converges to , for any there must be an such that for all there is:

    But now, the sequence of means in will be dominated by terms if is much larger than . So for a large enough , the mean drops below . This can be done with even smaller , so for , there is even:

    For there is now

    and we have convergence.

  2. No, the converse does not hold true. A counterexample is the sequence . It diverges (see the corresponding exercise above). However, for the Cesàro mean, there is

    This is obviously a null sequence.

  3. We apply Cauchy's limit theorem using . Since there is also .