Exercises: Series – Serlo

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Telescoping series[Bearbeiten]

Exercise

Investigate, whether the following series converge. If so, compute their limits.

Hint regarding subtask 3: We have . Why?

Hint regarding subtask 5: We have .

Solution

Subtask 1: This is a telescoping series with . Taking a look at the partial sums we get

As diverges, the series diverges as well.

Alternative solution: One can easily find a lower bound for the partial sum sequence:

As (harmonic series), is not bounded from above/below. Hence, the series diverges.

Subtask 2: We have


Obviously, this is a telescoping series with . We get:

Subtask 3: Take a look at the hint. We get

This is a more generalized version of a telescoping sum. The first and last two summands do not cancel:

Subtask 4: We have

We get the following telescoping series:

Subtask 5: Take a look at the hint! We get

Hence, we can calculate the series using two telescoping series:

Alternative solution: It holds that

Using the properties of telescoping series, we get:

Solution 6: It holds that

It follows that

Geometric series[Bearbeiten]

Exercise (Geometric series: Convergence and limits)

Investigate, whether the following series converge. If so, compute their limits.

  1. with for being even and for being odd.

Solution (Geometric series: Convergence and limits)

Subtask 1: We have

Subtask 2: As , this series diverges.

Subtask 3: The series converges. Using the computation rules for series, we get

Subtask 4: The series and converge. Using the computation rules for series, we get

Subtask 5: The series and converge. Using the computation rule for series, we get

Subtask 6: The series and converge. Using the computation rules for series, we get

Harmonic series[Bearbeiten]

Exercise (Harmonic series)

You may assume that converges and that holds.

  1. Explain why the series , and converge.
  2. Compute and .

Solution (Harmonic series)

Subtask 1:

1st series: The partial sum sequence is monotonously increasing as all summands are positive. Futhermore, is bounded from above as

Hence converges.

2nd series: We know that converges. Using the limit theorems for series, we get . Hence, this series converges.

3rd series: As the series converges absolutely, it converges.


Subtask 2:

1st series: We have

It follows that

2nd series: We have

Remark[Bearbeiten]

Analogously, for the generalizes harmonic series with we can show:


Exercise (Alternating harmonic series)

You may assume that converges and that holds.

Explain why the series converges and compute its limit.

Solution (Alternating harmonic series)

  • Convergence: We will show that the series converges absolutely. In the article about absolute convergence we have proven that this implies convergence. Let . As all summands are larger that zero, increases monotonically. Furthermore, we have
.

Hence, is bounded and converges.

  • Limit: We have

e-series[Bearbeiten]

Exercise (e-series)

Explain why the following series converge and compute their limits:

Solution (e-series)

Subtask 1: The partial sum sequence increases monotonously and is bounded from above as

Hence, the sequence converges.

Furthermore, we have

Alternative solution: Via telescoping sum. We have

Subtask 2: The partial sum sequence increases monotonously and is bounded from above as

Hence, the sequence converges.

Furthermore, we have

Rearrangement theorem for series[Bearbeiten]

Exercise (Rearrangement of alternating harmonic series)

The alternating harmonic series

and

converge to resp. . Show that the following rearrangements converge to the limits given.

Hint regarding subtask 2: Start off showing that with being the -th partial sum of the alternating harmonic series and being the -th partial sum of the rearranges series.

Solution (Rearrangement of alternating harmonic series)

Subtask 1: With and being the partial sums of the alternating harmonic series and the first rearrangement, we have:

converges and hence, tends to . Thus, converges and tends to as well.

Subtask 2 We have

As and tend to , tends to . We can conclude that converges and tends to .

Subtask 3: As , the series converges absolutely. Using the rearrangement theorem for absolutely convergent series, we get that every rearrangement of the series converges and tends to the same limit. Hence, the given rearrangement tends to .

Exercise (Rearrangement of convergent but not absolutely convergent series)

Prove the following statements: Let be a convergent -but not absolutely convergent - series. Then there exists a rearrangement of the series that...

  1. diverges but not to or .
  2. converges to an arbitrary .