Geometric series – Serlo

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Geometric series are series of the form . They are important within several proofs in real analysis. In particular, they are crucial for proving convergence or divergence of other series. We will derive some criteria using them, e.g. the ratio or the root criterion.

Geometric sum formula[Bearbeiten]

A video explaining the geometric series (in German).(YouTube-Video by the channel Quatematik)

We recall the geometric sum formula for partial sums of the geometric series. If you would like to know more about the geometric sum formula, take a look at the article „Geometrische Summenformel“ . The sum formula is proven there via induction. The proof of the sum formula reads as follows:

Theorem (Geometrische Summenformel)

For all real and for all there is:

Proof (Geometrische Summenformel)

We have

Geometric series[Bearbeiten]

A (German) song about the geometric series (Youtube-Video by DorFuchs)
The geometric series converges for , or .

We consider two cases: and .

Case [Bearbeiten]

We consider the geometric series for any , which especially means . The sum formula above applies to the partial sums in that case:

So the geometric series converges if and only if the sequence of partial sums converges. This is the case if and only if converges. We know that converges to if and only if and it converges to , if and only if . In this section, we only care about the first case of convergence:

If , then the geometric series converges.

Now, let us determine its limit:

Alternatively, convergence for can be shown directly, using the definition.

Exercise (alternative proof that the geometric series converges)

Prove that the geometric series with converges to .

How to get to the proof? (alternative proof that the geometric series converges)

We need to show that for each there is an , such that

for all

The geometric sum formula yields

Since the geometric sequence with converges to 0, so does . Hence, for every there is an with

for all

Since is constant, there is an with

for all

This implies the desired convergence.

Proof (alternative proof that the geometric series converges)

Let be given. The geometric sequence with converges to 0. Since , for a given there is an with

for all

With the geometric sum formula, this implies for all that

The limit of the series is hence: .

Case [Bearbeiten]

For , we have for all , that . Therefore, the sequence cannot converge to 0. So teh series must diverge (this argument is called term test and will be considered in detail, later)

The divergence becomes particularly obvious, if is positive, e.g. for . In this case, for all , we have and may estimate the partial sums: So the sequence of partial sums is bounded from below by the sequence , which in turn diverges to . So the series must diverge, as well.

Conclusion[Bearbeiten]

We have learned: for , and , the geometric series diverges. These three cases can be concluded into one case . However, if , then the geometric series converges to :

Theorem (geometric series)

The geometric series converges if and only if . In that case, the limit is , or written in shorthand notation:

Example (geometric series)

For , and there is

Example problems[Bearbeiten]

Problem 1[Bearbeiten]

Exercise (problems: geometric series)

Compute the values (=limits) of the following series:

Solution (problems: geometric series)

Solution sub-exercise 1:

Solution sub-exercise 2:

Solution sub-exercise 3:

Solution sub-exercise 4:

in a picture

Careful! The series starts with and not . So we need to shuffle terms around first before we can apply the sum formula with :

Solution sub-exercise 5:

This series starts with . We perform an index shift to get it starting with :

Problem 2[Bearbeiten]

Exercise (geometric series with special )

Let with and . Find the limits of the following six geometric series:

  1. and
  2. and
  3. and

Solution (geometric series with special )

Solution sub-exercise 1:

and

Solution sub-exercise 2:

and

Solution sub-exercise 3:

and

Problem 3[Bearbeiten]

Exercise (index shifting)

Let with . Find the limits of the three geometric series

  1. for

Solution (index shifting)

Solution sub-exercise 1:

Solution sub-exercise 2:

Solution sub-exercise 3:

For and there is

Problem 4[Bearbeiten]

Exercise (Sequences which relate to the geometric series)

Solve the following three problems:

  1. For all real and , prove the limit .
  2. For all with , prove the limit .
  3. Find the limits of the series and .

Solution (Sequences which relate to the geometric series)

Solution sub-exercise 1:

For all and we have to establish

the left-hand side is re-arranged as follows:

Solution sub-exercise 2:

In the chapter Beispiele von Grenzwerten , we proved that holds for (reason: grows exponentially and "beats" the linearly growing ). The limit theorems hence yield and . Therefore

Solution sub-exercise 3:

We re-use the solution to sub-exercise 2 with :

The second limit also follows from , using an index shift:

Solution (Alternative proof to sub-exercise 1)

We may also reconstruct the limit in a similar manner to the geometric sum formula: There is a factor of in the denominator, so applying the trick "multiply by and subtract" twice, we should get to the result. We start from

And multiply the equation by :

Next, both results are subtracted:

On the left side, we may factor out a second :

Solution (Alternative proof to sub-exercise 3)

We may also add and subtract a 1:

Hint

Analogously to sub-exercise 3, one may show for every that:

just replace by