Geometric series – Serlo

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

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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

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

We consider two cases:   and  .

Case  

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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  

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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

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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

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Problem 1

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Exercise (problems: geometric series)

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

  1.  
  2.  
  3.  
  4.  
  5.  

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

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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

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Exercise (index shifting)

Let   with  . Find the limits of the three geometric series

  1.  
  2.  
  3.   for  

Solution (index shifting)

Solution sub-exercise 1:

 

Solution sub-exercise 2:

 

Solution sub-exercise 3:

For   and   there is

 

Problem 4

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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