True or false: a sequence ${a_{n}}$ is convergent if and only if the subsequences $a_{2n}$ and $c_{2n + 1}$...












1












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This question already has an answer here:




  • Sufficiency to prove the convergence of a sequence using even and odd terms

    4 answers





True or false: a sequence ${a_{n}}$ is convergent if and only if the
subsequences $a_{2n}$ and $c_{2n + 1}$ both converge to the same
number




I think that the answer is true. I know that the converse is true because the subsequences of a convergent sequence always converge to the same number. I'm not sure about the forward direction though. I couldn't come up with a counterexample.










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marked as duplicate by Mike Earnest, RRL real-analysis
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Dec 14 '18 at 1:39


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    $c_{2n+1}$? Did you mean $a_{2n+1}$?
    $endgroup$
    – Alex R.
    Dec 14 '18 at 0:53










  • $begingroup$
    The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
    $endgroup$
    – JMoravitz
    Dec 14 '18 at 0:53










  • $begingroup$
    yes, i mean $a_{2n + 1}$
    $endgroup$
    – joseph
    Dec 14 '18 at 0:56










  • $begingroup$
    Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
    $endgroup$
    – William Sun
    Dec 14 '18 at 0:56










  • $begingroup$
    You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
    $endgroup$
    – Will M.
    Dec 14 '18 at 1:02
















1












$begingroup$



This question already has an answer here:




  • Sufficiency to prove the convergence of a sequence using even and odd terms

    4 answers





True or false: a sequence ${a_{n}}$ is convergent if and only if the
subsequences $a_{2n}$ and $c_{2n + 1}$ both converge to the same
number




I think that the answer is true. I know that the converse is true because the subsequences of a convergent sequence always converge to the same number. I'm not sure about the forward direction though. I couldn't come up with a counterexample.










share|cite|improve this question









$endgroup$



marked as duplicate by Mike Earnest, RRL real-analysis
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Dec 14 '18 at 1:39


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    $c_{2n+1}$? Did you mean $a_{2n+1}$?
    $endgroup$
    – Alex R.
    Dec 14 '18 at 0:53










  • $begingroup$
    The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
    $endgroup$
    – JMoravitz
    Dec 14 '18 at 0:53










  • $begingroup$
    yes, i mean $a_{2n + 1}$
    $endgroup$
    – joseph
    Dec 14 '18 at 0:56










  • $begingroup$
    Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
    $endgroup$
    – William Sun
    Dec 14 '18 at 0:56










  • $begingroup$
    You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
    $endgroup$
    – Will M.
    Dec 14 '18 at 1:02














1












1








1





$begingroup$



This question already has an answer here:




  • Sufficiency to prove the convergence of a sequence using even and odd terms

    4 answers





True or false: a sequence ${a_{n}}$ is convergent if and only if the
subsequences $a_{2n}$ and $c_{2n + 1}$ both converge to the same
number




I think that the answer is true. I know that the converse is true because the subsequences of a convergent sequence always converge to the same number. I'm not sure about the forward direction though. I couldn't come up with a counterexample.










share|cite|improve this question









$endgroup$





This question already has an answer here:




  • Sufficiency to prove the convergence of a sequence using even and odd terms

    4 answers





True or false: a sequence ${a_{n}}$ is convergent if and only if the
subsequences $a_{2n}$ and $c_{2n + 1}$ both converge to the same
number




I think that the answer is true. I know that the converse is true because the subsequences of a convergent sequence always converge to the same number. I'm not sure about the forward direction though. I couldn't come up with a counterexample.





This question already has an answer here:




  • Sufficiency to prove the convergence of a sequence using even and odd terms

    4 answers








real-analysis sequences-and-series






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share|cite|improve this question










asked Dec 14 '18 at 0:50









josephjoseph

500111




500111




marked as duplicate by Mike Earnest, RRL real-analysis
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Dec 14 '18 at 1:39


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Mike Earnest, RRL real-analysis
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Dec 14 '18 at 1:39


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    $c_{2n+1}$? Did you mean $a_{2n+1}$?
    $endgroup$
    – Alex R.
    Dec 14 '18 at 0:53










  • $begingroup$
    The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
    $endgroup$
    – JMoravitz
    Dec 14 '18 at 0:53










  • $begingroup$
    yes, i mean $a_{2n + 1}$
    $endgroup$
    – joseph
    Dec 14 '18 at 0:56










  • $begingroup$
    Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
    $endgroup$
    – William Sun
    Dec 14 '18 at 0:56










  • $begingroup$
    You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
    $endgroup$
    – Will M.
    Dec 14 '18 at 1:02


















  • $begingroup$
    $c_{2n+1}$? Did you mean $a_{2n+1}$?
    $endgroup$
    – Alex R.
    Dec 14 '18 at 0:53










  • $begingroup$
    The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
    $endgroup$
    – JMoravitz
    Dec 14 '18 at 0:53










  • $begingroup$
    yes, i mean $a_{2n + 1}$
    $endgroup$
    – joseph
    Dec 14 '18 at 0:56










  • $begingroup$
    Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
    $endgroup$
    – William Sun
    Dec 14 '18 at 0:56










  • $begingroup$
    You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
    $endgroup$
    – Will M.
    Dec 14 '18 at 1:02
















$begingroup$
$c_{2n+1}$? Did you mean $a_{2n+1}$?
$endgroup$
– Alex R.
Dec 14 '18 at 0:53




$begingroup$
$c_{2n+1}$? Did you mean $a_{2n+1}$?
$endgroup$
– Alex R.
Dec 14 '18 at 0:53












$begingroup$
The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
$endgroup$
– JMoravitz
Dec 14 '18 at 0:53




$begingroup$
The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
$endgroup$
– JMoravitz
Dec 14 '18 at 0:53












$begingroup$
yes, i mean $a_{2n + 1}$
$endgroup$
– joseph
Dec 14 '18 at 0:56




$begingroup$
yes, i mean $a_{2n + 1}$
$endgroup$
– joseph
Dec 14 '18 at 0:56












$begingroup$
Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
$endgroup$
– William Sun
Dec 14 '18 at 0:56




$begingroup$
Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
$endgroup$
– William Sun
Dec 14 '18 at 0:56












$begingroup$
You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
$endgroup$
– Will M.
Dec 14 '18 at 1:02




$begingroup$
You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
$endgroup$
– Will M.
Dec 14 '18 at 1:02










1 Answer
1






active

oldest

votes


















1












$begingroup$

Suppose



$$lim_{ntoinfty}a_{2n}=lim_{ntoinfty}a_{2n+1}=A$$



And let $;epsilon >0;$ be arbitrary, then there exist $;N_1,,N_2inBbb N;$ s.t.



$$begin{cases}|a_{2n}-A|<epsilon,,,,forall n>N_1\{}\|a_{2n+1}-A|<epsilon,,,,forall,n>N_2end{cases}$$



Let now $;M:=max{N_1,,N_2};$ , then for any $;n>M;$ , since $;n;$ is either even or odd....finish now with one line the proof.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Suppose



    $$lim_{ntoinfty}a_{2n}=lim_{ntoinfty}a_{2n+1}=A$$



    And let $;epsilon >0;$ be arbitrary, then there exist $;N_1,,N_2inBbb N;$ s.t.



    $$begin{cases}|a_{2n}-A|<epsilon,,,,forall n>N_1\{}\|a_{2n+1}-A|<epsilon,,,,forall,n>N_2end{cases}$$



    Let now $;M:=max{N_1,,N_2};$ , then for any $;n>M;$ , since $;n;$ is either even or odd....finish now with one line the proof.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Suppose



      $$lim_{ntoinfty}a_{2n}=lim_{ntoinfty}a_{2n+1}=A$$



      And let $;epsilon >0;$ be arbitrary, then there exist $;N_1,,N_2inBbb N;$ s.t.



      $$begin{cases}|a_{2n}-A|<epsilon,,,,forall n>N_1\{}\|a_{2n+1}-A|<epsilon,,,,forall,n>N_2end{cases}$$



      Let now $;M:=max{N_1,,N_2};$ , then for any $;n>M;$ , since $;n;$ is either even or odd....finish now with one line the proof.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Suppose



        $$lim_{ntoinfty}a_{2n}=lim_{ntoinfty}a_{2n+1}=A$$



        And let $;epsilon >0;$ be arbitrary, then there exist $;N_1,,N_2inBbb N;$ s.t.



        $$begin{cases}|a_{2n}-A|<epsilon,,,,forall n>N_1\{}\|a_{2n+1}-A|<epsilon,,,,forall,n>N_2end{cases}$$



        Let now $;M:=max{N_1,,N_2};$ , then for any $;n>M;$ , since $;n;$ is either even or odd....finish now with one line the proof.






        share|cite|improve this answer









        $endgroup$



        Suppose



        $$lim_{ntoinfty}a_{2n}=lim_{ntoinfty}a_{2n+1}=A$$



        And let $;epsilon >0;$ be arbitrary, then there exist $;N_1,,N_2inBbb N;$ s.t.



        $$begin{cases}|a_{2n}-A|<epsilon,,,,forall n>N_1\{}\|a_{2n+1}-A|<epsilon,,,,forall,n>N_2end{cases}$$



        Let now $;M:=max{N_1,,N_2};$ , then for any $;n>M;$ , since $;n;$ is either even or odd....finish now with one line the proof.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 1:07









        DonAntonioDonAntonio

        178k1493230




        178k1493230















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