Exercises: Subsequences, Accumulation points and Cauchy sequences – Serlo

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Exercises: Subsequences[Bearbeiten]

Exercise (Subsequences of the e-sequence)

  1. Calculate the limit
  2. Calculate the limit
  3. Calculate the limit using 1)

Solution (Subsequences of the e-sequence)

Subtask 1: As we know that the sequence has the limit , for the subsequence holds that:

Subtask 2: With

and being a subsequence of the sequence from 1, we get . Using the computation rules for limits, we get

Subtask 3: For holds that with the sequence from subtask 1.

As , is bounded from above, there exists an (actually ) with for all . Therefore, is an upper bound of . We even get

As , using the squeeze theorem we get

Exercises: Accumulation points[Bearbeiten]

Exercise (Accumulation points)

Determine all accumulation points of the sequences with and .

How to get to the proof? (Accumulation points)

  • Regarding : As converges to , all of its subsequences converge to . The sequence alternates between and . Hence, its accumulation points are and .
  • Regarding : For even -s, is a subsequence of the -sequence . For odd -s, is a subsequence of the sequence , which converges to .

Solution (Accumulation points)

  • As , one accumulation point of is .

Furthermore, we have . Hence, another accumulation point of is .

These are the only accumulation points of because every other subsequence of either diverges or converges to or .

  • We have , as . Thus, an accumulation point of is .

Furthermore, we have , as . Therefore, another accumulation point of is .

These are the only accumulation points of because every other subsequence of either diverges or converges to or .

Exercises: Lim inf and Lim sup [Bearbeiten]

Exercise (Lim inf and Lim sup)

Determine Lim inf and Lim sup of the sequences with and .

How to get to the proof? (Lim inf and Lim sup)

  • We start off investigating whether is bounded from above/below. This sequence is bounded from below but not from above. Hence, Lim inf equals the smallest accumulation point of . We can determine it looking at the subsequence including the odd sequence elements. As the sequence is not bounded from above, Lim sup equals .
  • As all of its subsequences are bounded, is bounded from above and below. In order to determine Lim sup and Lim inf of , we need to find the limits of its subsequences.

Solution (Lim inf and Lim sup)

  • As , we have . Hence, is bounded from below. then equals the smallest accumulation point of the sequence . For the subsequence holds that

Therefore, is an accumulation point of . There can't be other accumulation points, as all other subsequences of our sequence either converge to or diverge. Hence, we have .

Alternative solution: We have , as the odd sequence elements of monotonically decrease to . It follows that

With , we get

The archimedean property then states that there exists a with for all . Therefore, is not bounded from above and .

  • Regarding : We have

Hence, only has the accumulation points and . Furthermore, is bounded, as all subsequences are bounded as well. We get and .

Exercise (Lim inf and Lim sup 2)

  1. Let be a sequence of positive real numbers. Prove the following inequalities:
  2. Let be a sequence with

    Compute , , and .

  3. Derive the following statement from 1): Let be a sequence of positive real numbers with . Then we have .
  4. Prove the following limits:

Solution (Lim inf and Lim sup 2)

  1. Let be a positive sequence with

    We need to show the following inequalities:

    W.l.o.g. let and . By definition, is the smallest and the largest accumulation point of the sequence . Hence, for all () there exists an with

    Applying the inequality times setting , for an arbitrary with holds that:

    The expression in the middle is a telescoping product. Except for the numerator and the denominator , everything cancels out.

    Multiplying and taking the -th root leads to

    With , the term on the left side converges to and the one on the right side converges to for . Therefore, we get

    und

    As can be chosen arbitrarily small, we get

        and    

    As the inequality holds by definition, the claim follows.

    • For all we have , as the numerator and denominator are always positive. Hence, 0 is a lower bound of the quotient sequence . With being odd and even, it follows that
    for

    Hence, there exists a subsequence of the quotient sequence converging to zero. It follows that

    • With being even and being odd, we get
    for

    Hence, the quotient series doesn't have an upper bound and

    • With being odd, we have

    and being even leads to

    Hence, has the accumulation points and . It follows that

    and
  2. As , we get

    Using 1) and Der Sandwichsatz, we have

    It follows that .

  3. 1st limit: Let . It follows that

    Using part 3, we have .

    2nd limit: Let . It follows that

    Again using 3., we get .

Hint

Take a look at the inequalities proven in subtask 1. In subtask 2 even holds that

Hint

Taking a look at the limits from subtask 4 and using reciprocals, we get:

Exercises: Cauchy sequences[Bearbeiten]

Exercise (Cauchy criterion for sequences)

  1. Let and be a sequence of real numbers with for all . Show that the sequence converges.
  2. Let and . Show that the sequence converges and compute its limit.

Solution (Cauchy criterion for sequences)

Subtask 1: We will show that is a Cauchy sequence. The Cauchy criterion then states that converges. Using the assumptions made, we have:

If and , there exists a with . It follows that

From the inequalities follows that

From the archimedean axiom we can derive that for all there exists an such that for all holds

Hence, is a Cauchy sequence and therefore convergent.

Subtask 2: For the sequence holds true that

With , the criterion from subtask 1 is met. Hence, converges.

The limit can be computed as follows: Analog to subtask 1 - but without taking the absolute values -, we have

Applying the formula times leads to

Hence, we get

Using the computation rules for limits, we have