Exercises: Subsequences, Accumulation points and Cauchy sequences – Serlo

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