Basic Real Analysis - Two sub sequences converging to different limits












0












$begingroup$


Going through some real analysis questions and have hit a boundary, the question is



Consider the sequence:



$$a_n := (-1)^nbiggl( 1 + frac{1}{n^2}biggl) , n in mathbb{N}$$



(a) Find a subsequence $a_{n_k}$ which satifies $a_{n_k}$ $rightarrow$ 1 as n $rightarrow$ $infty$



(b) Find a subequence $a_{m_k}$ which satifies $a_{m_k}$ $rightarrow$ -1 as n $rightarrow$ $infty$



(c) Using results from (a) and (b) prove that the entire sequence cannot converge to a real number



My answer for (a) was $$a_{n_k} := biggl( 1 + frac{1}{n^2}biggl)$$
and (b) was $$a_{m_k} := biggl( -1 - frac{1}{n^2}biggl)$$
However, I am unsure of the proper way to go about proving (c) (other than just saying, two sub-sequences converge to different limits hence the series must be divergent). Any help would be greatly appreciated






real-analysis







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  • $begingroup$
    I don't see a particular reason you can't use that argument. Since the limit of $a_{n_k}$ is not equal to the limit of $a_{m_k}$, it must be divergent. This is because of a known result that if a sequence converges, then all its subsequences converge to the same limit. The contrapositive of this, and thus equivalent, is that if two subsequences converge to different limits, then the sequence itself cannot be convergent.
    $endgroup$
    – Eevee Trainer
    Nov 29 '18 at 2:42












  • $begingroup$
    Thanks, i guess i was overthinking a little. a conclusive sentence like that is pretty much what i was looking for
    $endgroup$
    – hellothere1
    Nov 29 '18 at 2:48






  • 1




    $begingroup$
    The use of k in your answer makes no sense.
    $endgroup$
    – William Elliot
    Nov 29 '18 at 2:52
















0












$begingroup$


Going through some real analysis questions and have hit a boundary, the question is



Consider the sequence:



$$a_n := (-1)^nbiggl( 1 + frac{1}{n^2}biggl) , n in mathbb{N}$$



(a) Find a subsequence $a_{n_k}$ which satifies $a_{n_k}$ $rightarrow$ 1 as n $rightarrow$ $infty$



(b) Find a subequence $a_{m_k}$ which satifies $a_{m_k}$ $rightarrow$ -1 as n $rightarrow$ $infty$



(c) Using results from (a) and (b) prove that the entire sequence cannot converge to a real number



My answer for (a) was $$a_{n_k} := biggl( 1 + frac{1}{n^2}biggl)$$
and (b) was $$a_{m_k} := biggl( -1 - frac{1}{n^2}biggl)$$
However, I am unsure of the proper way to go about proving (c) (other than just saying, two sub-sequences converge to different limits hence the series must be divergent). Any help would be greatly appreciated






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    I don't see a particular reason you can't use that argument. Since the limit of $a_{n_k}$ is not equal to the limit of $a_{m_k}$, it must be divergent. This is because of a known result that if a sequence converges, then all its subsequences converge to the same limit. The contrapositive of this, and thus equivalent, is that if two subsequences converge to different limits, then the sequence itself cannot be convergent.
    $endgroup$
    – Eevee Trainer
    Nov 29 '18 at 2:42












  • $begingroup$
    Thanks, i guess i was overthinking a little. a conclusive sentence like that is pretty much what i was looking for
    $endgroup$
    – hellothere1
    Nov 29 '18 at 2:48






  • 1




    $begingroup$
    The use of k in your answer makes no sense.
    $endgroup$
    – William Elliot
    Nov 29 '18 at 2:52














0












0








0





$begingroup$


Going through some real analysis questions and have hit a boundary, the question is



Consider the sequence:



$$a_n := (-1)^nbiggl( 1 + frac{1}{n^2}biggl) , n in mathbb{N}$$



(a) Find a subsequence $a_{n_k}$ which satifies $a_{n_k}$ $rightarrow$ 1 as n $rightarrow$ $infty$



(b) Find a subequence $a_{m_k}$ which satifies $a_{m_k}$ $rightarrow$ -1 as n $rightarrow$ $infty$



(c) Using results from (a) and (b) prove that the entire sequence cannot converge to a real number



My answer for (a) was $$a_{n_k} := biggl( 1 + frac{1}{n^2}biggl)$$
and (b) was $$a_{m_k} := biggl( -1 - frac{1}{n^2}biggl)$$
However, I am unsure of the proper way to go about proving (c) (other than just saying, two sub-sequences converge to different limits hence the series must be divergent). Any help would be greatly appreciated






real-analysis







share|cite|improve this question









$endgroup$




Going through some real analysis questions and have hit a boundary, the question is



Consider the sequence:



$$a_n := (-1)^nbiggl( 1 + frac{1}{n^2}biggl) , n in mathbb{N}$$



(a) Find a subsequence $a_{n_k}$ which satifies $a_{n_k}$ $rightarrow$ 1 as n $rightarrow$ $infty$



(b) Find a subequence $a_{m_k}$ which satifies $a_{m_k}$ $rightarrow$ -1 as n $rightarrow$ $infty$



(c) Using results from (a) and (b) prove that the entire sequence cannot converge to a real number



My answer for (a) was $$a_{n_k} := biggl( 1 + frac{1}{n^2}biggl)$$
and (b) was $$a_{m_k} := biggl( -1 - frac{1}{n^2}biggl)$$
However, I am unsure of the proper way to go about proving (c) (other than just saying, two sub-sequences converge to different limits hence the series must be divergent). Any help would be greatly appreciated






real-analysis




real-analysis






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 2:37









hellothere1hellothere1

101




101












  • $begingroup$
    I don't see a particular reason you can't use that argument. Since the limit of $a_{n_k}$ is not equal to the limit of $a_{m_k}$, it must be divergent. This is because of a known result that if a sequence converges, then all its subsequences converge to the same limit. The contrapositive of this, and thus equivalent, is that if two subsequences converge to different limits, then the sequence itself cannot be convergent.
    $endgroup$
    – Eevee Trainer
    Nov 29 '18 at 2:42












  • $begingroup$
    Thanks, i guess i was overthinking a little. a conclusive sentence like that is pretty much what i was looking for
    $endgroup$
    – hellothere1
    Nov 29 '18 at 2:48






  • 1




    $begingroup$
    The use of k in your answer makes no sense.
    $endgroup$
    – William Elliot
    Nov 29 '18 at 2:52


















  • $begingroup$
    I don't see a particular reason you can't use that argument. Since the limit of $a_{n_k}$ is not equal to the limit of $a_{m_k}$, it must be divergent. This is because of a known result that if a sequence converges, then all its subsequences converge to the same limit. The contrapositive of this, and thus equivalent, is that if two subsequences converge to different limits, then the sequence itself cannot be convergent.
    $endgroup$
    – Eevee Trainer
    Nov 29 '18 at 2:42












  • $begingroup$
    Thanks, i guess i was overthinking a little. a conclusive sentence like that is pretty much what i was looking for
    $endgroup$
    – hellothere1
    Nov 29 '18 at 2:48






  • 1




    $begingroup$
    The use of k in your answer makes no sense.
    $endgroup$
    – William Elliot
    Nov 29 '18 at 2:52
















$begingroup$
I don't see a particular reason you can't use that argument. Since the limit of $a_{n_k}$ is not equal to the limit of $a_{m_k}$, it must be divergent. This is because of a known result that if a sequence converges, then all its subsequences converge to the same limit. The contrapositive of this, and thus equivalent, is that if two subsequences converge to different limits, then the sequence itself cannot be convergent.
$endgroup$
– Eevee Trainer
Nov 29 '18 at 2:42






$begingroup$
I don't see a particular reason you can't use that argument. Since the limit of $a_{n_k}$ is not equal to the limit of $a_{m_k}$, it must be divergent. This is because of a known result that if a sequence converges, then all its subsequences converge to the same limit. The contrapositive of this, and thus equivalent, is that if two subsequences converge to different limits, then the sequence itself cannot be convergent.
$endgroup$
– Eevee Trainer
Nov 29 '18 at 2:42














$begingroup$
Thanks, i guess i was overthinking a little. a conclusive sentence like that is pretty much what i was looking for
$endgroup$
– hellothere1
Nov 29 '18 at 2:48




$begingroup$
Thanks, i guess i was overthinking a little. a conclusive sentence like that is pretty much what i was looking for
$endgroup$
– hellothere1
Nov 29 '18 at 2:48




1




1




$begingroup$
The use of k in your answer makes no sense.
$endgroup$
– William Elliot
Nov 29 '18 at 2:52




$begingroup$
The use of k in your answer makes no sense.
$endgroup$
– William Elliot
Nov 29 '18 at 2:52










1 Answer
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$begingroup$

I think you're on the right track. An approach to this is to make use of the fact that any limit that a sequence converges to must be unique.



Since you have shown in the previous two parts that there are two subsequences that converge to two different limits, I do think that you can automatically conclude that this immediately voids the possibility of the sequence converging.



However, I do think that it may be possible to come up with a 'proof' of sorts (by contradiction):



Assume that the sequence $a_n := (-1)^n (1+frac{1}{n^2})$ converges to a limit, say $a$. This implies that the sequence is a Cauchy sequence. By the definition of a Cauchy sequence, we can set $epsilon = 1$, and make the claim that there exists some $kin mathbb{N}$ such that $forall m,n > k$, that $|a_m - a_n| < epsilon = 1$



So all that is left, is to show that the claim above is false. It is straightforward to do so and we will do this by construction - select $m = 2k, n = 2k+1$, then $a_m$ belongs to the first subsequence that converges to 1, and $a_n$ belongs to the second subsequence that converges to -1.



For completeness:
$$
begin{align}
|a_m - a_n| &= Big|(1 + frac{1}{4k^2}) - (- 1 - frac{1}{(2k+1)^2})Big| \
&= Big|2 + frac{1}{4k^2} + frac{1}{(2k+1)^2}Big| \
&> 2 \
&> 1 = epsilon quad text{(contradiction)}
end{align}
$$
Thus, what we have shown is that the sequence cannot converge, since it is not a Cauchy sequence.






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    0












    $begingroup$

    I think you're on the right track. An approach to this is to make use of the fact that any limit that a sequence converges to must be unique.



    Since you have shown in the previous two parts that there are two subsequences that converge to two different limits, I do think that you can automatically conclude that this immediately voids the possibility of the sequence converging.



    However, I do think that it may be possible to come up with a 'proof' of sorts (by contradiction):



    Assume that the sequence $a_n := (-1)^n (1+frac{1}{n^2})$ converges to a limit, say $a$. This implies that the sequence is a Cauchy sequence. By the definition of a Cauchy sequence, we can set $epsilon = 1$, and make the claim that there exists some $kin mathbb{N}$ such that $forall m,n > k$, that $|a_m - a_n| < epsilon = 1$



    So all that is left, is to show that the claim above is false. It is straightforward to do so and we will do this by construction - select $m = 2k, n = 2k+1$, then $a_m$ belongs to the first subsequence that converges to 1, and $a_n$ belongs to the second subsequence that converges to -1.



    For completeness:
    $$
    begin{align}
    |a_m - a_n| &= Big|(1 + frac{1}{4k^2}) - (- 1 - frac{1}{(2k+1)^2})Big| \
    &= Big|2 + frac{1}{4k^2} + frac{1}{(2k+1)^2}Big| \
    &> 2 \
    &> 1 = epsilon quad text{(contradiction)}
    end{align}
    $$
    Thus, what we have shown is that the sequence cannot converge, since it is not a Cauchy sequence.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I think you're on the right track. An approach to this is to make use of the fact that any limit that a sequence converges to must be unique.



      Since you have shown in the previous two parts that there are two subsequences that converge to two different limits, I do think that you can automatically conclude that this immediately voids the possibility of the sequence converging.



      However, I do think that it may be possible to come up with a 'proof' of sorts (by contradiction):



      Assume that the sequence $a_n := (-1)^n (1+frac{1}{n^2})$ converges to a limit, say $a$. This implies that the sequence is a Cauchy sequence. By the definition of a Cauchy sequence, we can set $epsilon = 1$, and make the claim that there exists some $kin mathbb{N}$ such that $forall m,n > k$, that $|a_m - a_n| < epsilon = 1$



      So all that is left, is to show that the claim above is false. It is straightforward to do so and we will do this by construction - select $m = 2k, n = 2k+1$, then $a_m$ belongs to the first subsequence that converges to 1, and $a_n$ belongs to the second subsequence that converges to -1.



      For completeness:
      $$
      begin{align}
      |a_m - a_n| &= Big|(1 + frac{1}{4k^2}) - (- 1 - frac{1}{(2k+1)^2})Big| \
      &= Big|2 + frac{1}{4k^2} + frac{1}{(2k+1)^2}Big| \
      &> 2 \
      &> 1 = epsilon quad text{(contradiction)}
      end{align}
      $$
      Thus, what we have shown is that the sequence cannot converge, since it is not a Cauchy sequence.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I think you're on the right track. An approach to this is to make use of the fact that any limit that a sequence converges to must be unique.



        Since you have shown in the previous two parts that there are two subsequences that converge to two different limits, I do think that you can automatically conclude that this immediately voids the possibility of the sequence converging.



        However, I do think that it may be possible to come up with a 'proof' of sorts (by contradiction):



        Assume that the sequence $a_n := (-1)^n (1+frac{1}{n^2})$ converges to a limit, say $a$. This implies that the sequence is a Cauchy sequence. By the definition of a Cauchy sequence, we can set $epsilon = 1$, and make the claim that there exists some $kin mathbb{N}$ such that $forall m,n > k$, that $|a_m - a_n| < epsilon = 1$



        So all that is left, is to show that the claim above is false. It is straightforward to do so and we will do this by construction - select $m = 2k, n = 2k+1$, then $a_m$ belongs to the first subsequence that converges to 1, and $a_n$ belongs to the second subsequence that converges to -1.



        For completeness:
        $$
        begin{align}
        |a_m - a_n| &= Big|(1 + frac{1}{4k^2}) - (- 1 - frac{1}{(2k+1)^2})Big| \
        &= Big|2 + frac{1}{4k^2} + frac{1}{(2k+1)^2}Big| \
        &> 2 \
        &> 1 = epsilon quad text{(contradiction)}
        end{align}
        $$
        Thus, what we have shown is that the sequence cannot converge, since it is not a Cauchy sequence.






        share|cite|improve this answer









        $endgroup$



        I think you're on the right track. An approach to this is to make use of the fact that any limit that a sequence converges to must be unique.



        Since you have shown in the previous two parts that there are two subsequences that converge to two different limits, I do think that you can automatically conclude that this immediately voids the possibility of the sequence converging.



        However, I do think that it may be possible to come up with a 'proof' of sorts (by contradiction):



        Assume that the sequence $a_n := (-1)^n (1+frac{1}{n^2})$ converges to a limit, say $a$. This implies that the sequence is a Cauchy sequence. By the definition of a Cauchy sequence, we can set $epsilon = 1$, and make the claim that there exists some $kin mathbb{N}$ such that $forall m,n > k$, that $|a_m - a_n| < epsilon = 1$



        So all that is left, is to show that the claim above is false. It is straightforward to do so and we will do this by construction - select $m = 2k, n = 2k+1$, then $a_m$ belongs to the first subsequence that converges to 1, and $a_n$ belongs to the second subsequence that converges to -1.



        For completeness:
        $$
        begin{align}
        |a_m - a_n| &= Big|(1 + frac{1}{4k^2}) - (- 1 - frac{1}{(2k+1)^2})Big| \
        &= Big|2 + frac{1}{4k^2} + frac{1}{(2k+1)^2}Big| \
        &> 2 \
        &> 1 = epsilon quad text{(contradiction)}
        end{align}
        $$
        Thus, what we have shown is that the sequence cannot converge, since it is not a Cauchy sequence.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 2:58









        Sean LeeSean Lee

        383111




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