question on subsequence and convergence












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We are given the sequence $a_n$ and it's subsequences, $a_{2n}$, $a_{2n+1}$ and $a_{3n}$, which are each convergent(and each converge to the same limit). I know that we can use $a_{2n}$ and $a_{2n+1}$ to show that $a_n$ is convergent , but can one deduce the same (that $a_n$ is convergent) with just $a_{2n}$ and $a_{3n}$ ? I know that method is similar to $a_{2n}$ and $a_{2n+1}$ but $2n$ & $3n$ do not account for all the terms.










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  • Consider the sequence in which $a_k=0$ whenever $k$ is divisible by $2$ or by $3$ but $a_k=1$ for all the other values of $k$ (those that are $equivpm1pmod6$).
    – Andreas Blass
    Dec 27 '18 at 1:23










  • "but 2n & 3n do not account for all the terms. " Which means you can not conclude anything. End of story. (Well, actually they don't have to account for all terms but the must account for all be a finite number of terms. If there are an infinite number of terms that are unaccounted for, those terms can do whatever the #### they want and need not converge. ... Now end of story.)
    – fleablood
    Dec 27 '18 at 2:28












  • Thanks for the help guys!
    – forward_behind1
    Dec 27 '18 at 15:41
















0














We are given the sequence $a_n$ and it's subsequences, $a_{2n}$, $a_{2n+1}$ and $a_{3n}$, which are each convergent(and each converge to the same limit). I know that we can use $a_{2n}$ and $a_{2n+1}$ to show that $a_n$ is convergent , but can one deduce the same (that $a_n$ is convergent) with just $a_{2n}$ and $a_{3n}$ ? I know that method is similar to $a_{2n}$ and $a_{2n+1}$ but $2n$ & $3n$ do not account for all the terms.










share|cite|improve this question







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  • Consider the sequence in which $a_k=0$ whenever $k$ is divisible by $2$ or by $3$ but $a_k=1$ for all the other values of $k$ (those that are $equivpm1pmod6$).
    – Andreas Blass
    Dec 27 '18 at 1:23










  • "but 2n & 3n do not account for all the terms. " Which means you can not conclude anything. End of story. (Well, actually they don't have to account for all terms but the must account for all be a finite number of terms. If there are an infinite number of terms that are unaccounted for, those terms can do whatever the #### they want and need not converge. ... Now end of story.)
    – fleablood
    Dec 27 '18 at 2:28












  • Thanks for the help guys!
    – forward_behind1
    Dec 27 '18 at 15:41














0












0








0







We are given the sequence $a_n$ and it's subsequences, $a_{2n}$, $a_{2n+1}$ and $a_{3n}$, which are each convergent(and each converge to the same limit). I know that we can use $a_{2n}$ and $a_{2n+1}$ to show that $a_n$ is convergent , but can one deduce the same (that $a_n$ is convergent) with just $a_{2n}$ and $a_{3n}$ ? I know that method is similar to $a_{2n}$ and $a_{2n+1}$ but $2n$ & $3n$ do not account for all the terms.










share|cite|improve this question







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We are given the sequence $a_n$ and it's subsequences, $a_{2n}$, $a_{2n+1}$ and $a_{3n}$, which are each convergent(and each converge to the same limit). I know that we can use $a_{2n}$ and $a_{2n+1}$ to show that $a_n$ is convergent , but can one deduce the same (that $a_n$ is convergent) with just $a_{2n}$ and $a_{3n}$ ? I know that method is similar to $a_{2n}$ and $a_{2n+1}$ but $2n$ & $3n$ do not account for all the terms.







sequences-and-series






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asked Dec 27 '18 at 1:19









forward_behind1

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32




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  • Consider the sequence in which $a_k=0$ whenever $k$ is divisible by $2$ or by $3$ but $a_k=1$ for all the other values of $k$ (those that are $equivpm1pmod6$).
    – Andreas Blass
    Dec 27 '18 at 1:23










  • "but 2n & 3n do not account for all the terms. " Which means you can not conclude anything. End of story. (Well, actually they don't have to account for all terms but the must account for all be a finite number of terms. If there are an infinite number of terms that are unaccounted for, those terms can do whatever the #### they want and need not converge. ... Now end of story.)
    – fleablood
    Dec 27 '18 at 2:28












  • Thanks for the help guys!
    – forward_behind1
    Dec 27 '18 at 15:41


















  • Consider the sequence in which $a_k=0$ whenever $k$ is divisible by $2$ or by $3$ but $a_k=1$ for all the other values of $k$ (those that are $equivpm1pmod6$).
    – Andreas Blass
    Dec 27 '18 at 1:23










  • "but 2n & 3n do not account for all the terms. " Which means you can not conclude anything. End of story. (Well, actually they don't have to account for all terms but the must account for all be a finite number of terms. If there are an infinite number of terms that are unaccounted for, those terms can do whatever the #### they want and need not converge. ... Now end of story.)
    – fleablood
    Dec 27 '18 at 2:28












  • Thanks for the help guys!
    – forward_behind1
    Dec 27 '18 at 15:41
















Consider the sequence in which $a_k=0$ whenever $k$ is divisible by $2$ or by $3$ but $a_k=1$ for all the other values of $k$ (those that are $equivpm1pmod6$).
– Andreas Blass
Dec 27 '18 at 1:23




Consider the sequence in which $a_k=0$ whenever $k$ is divisible by $2$ or by $3$ but $a_k=1$ for all the other values of $k$ (those that are $equivpm1pmod6$).
– Andreas Blass
Dec 27 '18 at 1:23












"but 2n & 3n do not account for all the terms. " Which means you can not conclude anything. End of story. (Well, actually they don't have to account for all terms but the must account for all be a finite number of terms. If there are an infinite number of terms that are unaccounted for, those terms can do whatever the #### they want and need not converge. ... Now end of story.)
– fleablood
Dec 27 '18 at 2:28






"but 2n & 3n do not account for all the terms. " Which means you can not conclude anything. End of story. (Well, actually they don't have to account for all terms but the must account for all be a finite number of terms. If there are an infinite number of terms that are unaccounted for, those terms can do whatever the #### they want and need not converge. ... Now end of story.)
– fleablood
Dec 27 '18 at 2:28














Thanks for the help guys!
– forward_behind1
Dec 27 '18 at 15:41




Thanks for the help guys!
– forward_behind1
Dec 27 '18 at 15:41










4 Answers
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I know that method is similar to $a_{2n} and $a_{2n+1} but $2_n$ & $3_n$ do not account for all the terms.




ANd that makes ALL the difference! If there are terms unaccounted for, those terms can do anything they want!



The point is that all $a_k$ that $k$ is either of the form $2n$ or $2n +1$ so $a_k$ in one of the sequences or the other. And if they both converge to the same limit then then whole sequence converges to the same limit.



And if you had any set of sequences $a_{k_i}, a_{j_i}, a_{m_i}$ and if the $k_i, j_i, m_i$ partitioned all the positive integers, or partitioned all but a finite number of the the positive integers so that at some point any $a_n$ must belong to one of the sequences everything would be fine.



But if there are always terms that are not in either sequence, it need not converge. $2n$ and $3n$ in no way cover all possible positive integers so we can conclude nothing from $a_{2n}$ and $a_{3n}$ converging.



Example let $a_k$ be defined as $a_k = 0$ if $2|k$ or if $3|k$ but let $a_k = k$ if neither $2$ nor $3$ divide $k$.



Then all $a_{2n} = 0$ and all $a_{3n} = 0$ so $a_{2n} to 0$ and $a_{3n}to 0$ but clearly $a_n notto 0$. (Because there are infinitely many $n$ not divisible by $2$ or by $3$ and those terms do not converge to anything.)



======



I suspect the question actually was if $a_{2n} to d$ and $a_{3n} to c$ and $a_{2n+1} to b$ then prove $a_n$ converges and determine what it converges to.



1) As $a_{6n} subset a_{2n}$ and $a_{6n} subset a_{3n}$ then $a_{6n}to d$ and $a_{6n} to c$ and $d=c$.



2) As $a_{6n + 3} subset a_{3n}$ and $a_{6n+3} subset a_{2n+1}$ then $a_{6n+3} to c=d$ and $a_{6n+3}to b$ and $b = c =d$.



3) As $a_{2n+1}$ and $a_{2n}$ completely partition $a_n$, then all $a_k$ are in one sequence or the other and those the sequence will converge to $b=c=d$.






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    0














    The comment from Andreas Blass contains the answer. The only thing one need to answer your question is a counter-example. Take for instance the series $(u_n)$ such that:




    • for all prime number $p>3$, $u_p = 1$

    • for all other integer $k$, $u_k = 0$


    The Euclid's theorem asserts that there are an infinity of prime numbers, therefore the series can not converge even though any terms of $u_{2n}$ or $u_{3n}$ will be $0$.






    share|cite|improve this answer





























      0














      I think the correct question is : $(a_{2n})$, $(a_{2n+1})$ and $(a_{3n})$ are convergent, but you don't have the additional information that the limits are the same.



      Now prove that $(a_n)$ is convergent.






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      • Um... you can't. But you can prove that the limits are all the same.
        – fleablood
        Dec 27 '18 at 2:26










      • Yes you can : knowing that $(a_{2n})$ and $(a_{2n+1})$ both are convergent isn't enough (think of $(-1)^n$). But if you know that $(a_{3n})$ is also convergent, then you can prove that $(a_{2n})$ and $(a_{2n+1})$ have the same limit. Now THAT is enough to prove that $(a_n)$ is convergent.
        – Nicolas FRANCOIS
        Dec 28 '18 at 15:29










      • Oh, I'm not sure how I misread you answer but I thought you hadn't including the $a_{2n+1}$ but just had the $a_{2n}$ and $a_{3n}$ as was the OP's question. As there are infinitely many terms belonging to both sequences they most converge to same. But, without, $a_{2n+1}$ there are infinitely many in neither so one couldn't conclude $a_n$ converge.
        – fleablood
        Dec 28 '18 at 16:47



















      0














      The idea behind the proof of $ a_2n $ and $ a_(2n+1) $ being sufficient to conclude about the limit of $ a_n $ is that sequences of indices $ 2n $ and $ (2n+1) $ partition the family of indices $ I_n $, thus complete "the puzzle" if you wish. ie. they generate the family of indices $ I_n $, a subgroup of $ mathbb{Z}^+ $.



      So, the procedure is without alternative; and the question about limit can be reduced to prove that **$<2n+1,3n>$ is also a generator for $ mathbb{Z}^+ $. To generate means that, there exists $ k_1,k_2 in mathbb{Z} $ such that:



      $$ (1) iff k_1 (2n+1) + k_2 (3n) = N $$



      where $ N $ is just any integer possible.



      Observe that,



      $$(1) iff n (2 k_1) + k_1 + n (3 k_2) = N $$



      $$(1) iff n (2k_1 + 3k_2) + k_1 = N $$



      We introduce: $ k_3 = 2k_1 + 3k_2 $:



      $$(1) iff n k_3 + k_1 = N $$



      In fact, we conclude that, for any $ k_3 $: $ <2n+1,3n> $ generates the quotient group $ mathbb{Z}/mathbb{Z}_(k_3) $. And, as we have no criteria over $k_3$, these two subsequences above are not sufficient to conclude anything about the main sequence's limit.






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        I know that method is similar to $a_{2n} and $a_{2n+1} but $2_n$ & $3_n$ do not account for all the terms.




        ANd that makes ALL the difference! If there are terms unaccounted for, those terms can do anything they want!



        The point is that all $a_k$ that $k$ is either of the form $2n$ or $2n +1$ so $a_k$ in one of the sequences or the other. And if they both converge to the same limit then then whole sequence converges to the same limit.



        And if you had any set of sequences $a_{k_i}, a_{j_i}, a_{m_i}$ and if the $k_i, j_i, m_i$ partitioned all the positive integers, or partitioned all but a finite number of the the positive integers so that at some point any $a_n$ must belong to one of the sequences everything would be fine.



        But if there are always terms that are not in either sequence, it need not converge. $2n$ and $3n$ in no way cover all possible positive integers so we can conclude nothing from $a_{2n}$ and $a_{3n}$ converging.



        Example let $a_k$ be defined as $a_k = 0$ if $2|k$ or if $3|k$ but let $a_k = k$ if neither $2$ nor $3$ divide $k$.



        Then all $a_{2n} = 0$ and all $a_{3n} = 0$ so $a_{2n} to 0$ and $a_{3n}to 0$ but clearly $a_n notto 0$. (Because there are infinitely many $n$ not divisible by $2$ or by $3$ and those terms do not converge to anything.)



        ======



        I suspect the question actually was if $a_{2n} to d$ and $a_{3n} to c$ and $a_{2n+1} to b$ then prove $a_n$ converges and determine what it converges to.



        1) As $a_{6n} subset a_{2n}$ and $a_{6n} subset a_{3n}$ then $a_{6n}to d$ and $a_{6n} to c$ and $d=c$.



        2) As $a_{6n + 3} subset a_{3n}$ and $a_{6n+3} subset a_{2n+1}$ then $a_{6n+3} to c=d$ and $a_{6n+3}to b$ and $b = c =d$.



        3) As $a_{2n+1}$ and $a_{2n}$ completely partition $a_n$, then all $a_k$ are in one sequence or the other and those the sequence will converge to $b=c=d$.






        share|cite|improve this answer




























          0















          I know that method is similar to $a_{2n} and $a_{2n+1} but $2_n$ & $3_n$ do not account for all the terms.




          ANd that makes ALL the difference! If there are terms unaccounted for, those terms can do anything they want!



          The point is that all $a_k$ that $k$ is either of the form $2n$ or $2n +1$ so $a_k$ in one of the sequences or the other. And if they both converge to the same limit then then whole sequence converges to the same limit.



          And if you had any set of sequences $a_{k_i}, a_{j_i}, a_{m_i}$ and if the $k_i, j_i, m_i$ partitioned all the positive integers, or partitioned all but a finite number of the the positive integers so that at some point any $a_n$ must belong to one of the sequences everything would be fine.



          But if there are always terms that are not in either sequence, it need not converge. $2n$ and $3n$ in no way cover all possible positive integers so we can conclude nothing from $a_{2n}$ and $a_{3n}$ converging.



          Example let $a_k$ be defined as $a_k = 0$ if $2|k$ or if $3|k$ but let $a_k = k$ if neither $2$ nor $3$ divide $k$.



          Then all $a_{2n} = 0$ and all $a_{3n} = 0$ so $a_{2n} to 0$ and $a_{3n}to 0$ but clearly $a_n notto 0$. (Because there are infinitely many $n$ not divisible by $2$ or by $3$ and those terms do not converge to anything.)



          ======



          I suspect the question actually was if $a_{2n} to d$ and $a_{3n} to c$ and $a_{2n+1} to b$ then prove $a_n$ converges and determine what it converges to.



          1) As $a_{6n} subset a_{2n}$ and $a_{6n} subset a_{3n}$ then $a_{6n}to d$ and $a_{6n} to c$ and $d=c$.



          2) As $a_{6n + 3} subset a_{3n}$ and $a_{6n+3} subset a_{2n+1}$ then $a_{6n+3} to c=d$ and $a_{6n+3}to b$ and $b = c =d$.



          3) As $a_{2n+1}$ and $a_{2n}$ completely partition $a_n$, then all $a_k$ are in one sequence or the other and those the sequence will converge to $b=c=d$.






          share|cite|improve this answer


























            0












            0








            0







            I know that method is similar to $a_{2n} and $a_{2n+1} but $2_n$ & $3_n$ do not account for all the terms.




            ANd that makes ALL the difference! If there are terms unaccounted for, those terms can do anything they want!



            The point is that all $a_k$ that $k$ is either of the form $2n$ or $2n +1$ so $a_k$ in one of the sequences or the other. And if they both converge to the same limit then then whole sequence converges to the same limit.



            And if you had any set of sequences $a_{k_i}, a_{j_i}, a_{m_i}$ and if the $k_i, j_i, m_i$ partitioned all the positive integers, or partitioned all but a finite number of the the positive integers so that at some point any $a_n$ must belong to one of the sequences everything would be fine.



            But if there are always terms that are not in either sequence, it need not converge. $2n$ and $3n$ in no way cover all possible positive integers so we can conclude nothing from $a_{2n}$ and $a_{3n}$ converging.



            Example let $a_k$ be defined as $a_k = 0$ if $2|k$ or if $3|k$ but let $a_k = k$ if neither $2$ nor $3$ divide $k$.



            Then all $a_{2n} = 0$ and all $a_{3n} = 0$ so $a_{2n} to 0$ and $a_{3n}to 0$ but clearly $a_n notto 0$. (Because there are infinitely many $n$ not divisible by $2$ or by $3$ and those terms do not converge to anything.)



            ======



            I suspect the question actually was if $a_{2n} to d$ and $a_{3n} to c$ and $a_{2n+1} to b$ then prove $a_n$ converges and determine what it converges to.



            1) As $a_{6n} subset a_{2n}$ and $a_{6n} subset a_{3n}$ then $a_{6n}to d$ and $a_{6n} to c$ and $d=c$.



            2) As $a_{6n + 3} subset a_{3n}$ and $a_{6n+3} subset a_{2n+1}$ then $a_{6n+3} to c=d$ and $a_{6n+3}to b$ and $b = c =d$.



            3) As $a_{2n+1}$ and $a_{2n}$ completely partition $a_n$, then all $a_k$ are in one sequence or the other and those the sequence will converge to $b=c=d$.






            share|cite|improve this answer















            I know that method is similar to $a_{2n} and $a_{2n+1} but $2_n$ & $3_n$ do not account for all the terms.




            ANd that makes ALL the difference! If there are terms unaccounted for, those terms can do anything they want!



            The point is that all $a_k$ that $k$ is either of the form $2n$ or $2n +1$ so $a_k$ in one of the sequences or the other. And if they both converge to the same limit then then whole sequence converges to the same limit.



            And if you had any set of sequences $a_{k_i}, a_{j_i}, a_{m_i}$ and if the $k_i, j_i, m_i$ partitioned all the positive integers, or partitioned all but a finite number of the the positive integers so that at some point any $a_n$ must belong to one of the sequences everything would be fine.



            But if there are always terms that are not in either sequence, it need not converge. $2n$ and $3n$ in no way cover all possible positive integers so we can conclude nothing from $a_{2n}$ and $a_{3n}$ converging.



            Example let $a_k$ be defined as $a_k = 0$ if $2|k$ or if $3|k$ but let $a_k = k$ if neither $2$ nor $3$ divide $k$.



            Then all $a_{2n} = 0$ and all $a_{3n} = 0$ so $a_{2n} to 0$ and $a_{3n}to 0$ but clearly $a_n notto 0$. (Because there are infinitely many $n$ not divisible by $2$ or by $3$ and those terms do not converge to anything.)



            ======



            I suspect the question actually was if $a_{2n} to d$ and $a_{3n} to c$ and $a_{2n+1} to b$ then prove $a_n$ converges and determine what it converges to.



            1) As $a_{6n} subset a_{2n}$ and $a_{6n} subset a_{3n}$ then $a_{6n}to d$ and $a_{6n} to c$ and $d=c$.



            2) As $a_{6n + 3} subset a_{3n}$ and $a_{6n+3} subset a_{2n+1}$ then $a_{6n+3} to c=d$ and $a_{6n+3}to b$ and $b = c =d$.



            3) As $a_{2n+1}$ and $a_{2n}$ completely partition $a_n$, then all $a_k$ are in one sequence or the other and those the sequence will converge to $b=c=d$.







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            edited Dec 28 '18 at 17:04

























            answered Dec 27 '18 at 2:24









            fleablood

            68.2k22685




            68.2k22685























                0














                The comment from Andreas Blass contains the answer. The only thing one need to answer your question is a counter-example. Take for instance the series $(u_n)$ such that:




                • for all prime number $p>3$, $u_p = 1$

                • for all other integer $k$, $u_k = 0$


                The Euclid's theorem asserts that there are an infinity of prime numbers, therefore the series can not converge even though any terms of $u_{2n}$ or $u_{3n}$ will be $0$.






                share|cite|improve this answer


























                  0














                  The comment from Andreas Blass contains the answer. The only thing one need to answer your question is a counter-example. Take for instance the series $(u_n)$ such that:




                  • for all prime number $p>3$, $u_p = 1$

                  • for all other integer $k$, $u_k = 0$


                  The Euclid's theorem asserts that there are an infinity of prime numbers, therefore the series can not converge even though any terms of $u_{2n}$ or $u_{3n}$ will be $0$.






                  share|cite|improve this answer
























                    0












                    0








                    0






                    The comment from Andreas Blass contains the answer. The only thing one need to answer your question is a counter-example. Take for instance the series $(u_n)$ such that:




                    • for all prime number $p>3$, $u_p = 1$

                    • for all other integer $k$, $u_k = 0$


                    The Euclid's theorem asserts that there are an infinity of prime numbers, therefore the series can not converge even though any terms of $u_{2n}$ or $u_{3n}$ will be $0$.






                    share|cite|improve this answer












                    The comment from Andreas Blass contains the answer. The only thing one need to answer your question is a counter-example. Take for instance the series $(u_n)$ such that:




                    • for all prime number $p>3$, $u_p = 1$

                    • for all other integer $k$, $u_k = 0$


                    The Euclid's theorem asserts that there are an infinity of prime numbers, therefore the series can not converge even though any terms of $u_{2n}$ or $u_{3n}$ will be $0$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 27 '18 at 1:35









                    TryingToGetOut

                    488




                    488























                        0














                        I think the correct question is : $(a_{2n})$, $(a_{2n+1})$ and $(a_{3n})$ are convergent, but you don't have the additional information that the limits are the same.



                        Now prove that $(a_n)$ is convergent.






                        share|cite|improve this answer





















                        • Um... you can't. But you can prove that the limits are all the same.
                          – fleablood
                          Dec 27 '18 at 2:26










                        • Yes you can : knowing that $(a_{2n})$ and $(a_{2n+1})$ both are convergent isn't enough (think of $(-1)^n$). But if you know that $(a_{3n})$ is also convergent, then you can prove that $(a_{2n})$ and $(a_{2n+1})$ have the same limit. Now THAT is enough to prove that $(a_n)$ is convergent.
                          – Nicolas FRANCOIS
                          Dec 28 '18 at 15:29










                        • Oh, I'm not sure how I misread you answer but I thought you hadn't including the $a_{2n+1}$ but just had the $a_{2n}$ and $a_{3n}$ as was the OP's question. As there are infinitely many terms belonging to both sequences they most converge to same. But, without, $a_{2n+1}$ there are infinitely many in neither so one couldn't conclude $a_n$ converge.
                          – fleablood
                          Dec 28 '18 at 16:47
















                        0














                        I think the correct question is : $(a_{2n})$, $(a_{2n+1})$ and $(a_{3n})$ are convergent, but you don't have the additional information that the limits are the same.



                        Now prove that $(a_n)$ is convergent.






                        share|cite|improve this answer





















                        • Um... you can't. But you can prove that the limits are all the same.
                          – fleablood
                          Dec 27 '18 at 2:26










                        • Yes you can : knowing that $(a_{2n})$ and $(a_{2n+1})$ both are convergent isn't enough (think of $(-1)^n$). But if you know that $(a_{3n})$ is also convergent, then you can prove that $(a_{2n})$ and $(a_{2n+1})$ have the same limit. Now THAT is enough to prove that $(a_n)$ is convergent.
                          – Nicolas FRANCOIS
                          Dec 28 '18 at 15:29










                        • Oh, I'm not sure how I misread you answer but I thought you hadn't including the $a_{2n+1}$ but just had the $a_{2n}$ and $a_{3n}$ as was the OP's question. As there are infinitely many terms belonging to both sequences they most converge to same. But, without, $a_{2n+1}$ there are infinitely many in neither so one couldn't conclude $a_n$ converge.
                          – fleablood
                          Dec 28 '18 at 16:47














                        0












                        0








                        0






                        I think the correct question is : $(a_{2n})$, $(a_{2n+1})$ and $(a_{3n})$ are convergent, but you don't have the additional information that the limits are the same.



                        Now prove that $(a_n)$ is convergent.






                        share|cite|improve this answer












                        I think the correct question is : $(a_{2n})$, $(a_{2n+1})$ and $(a_{3n})$ are convergent, but you don't have the additional information that the limits are the same.



                        Now prove that $(a_n)$ is convergent.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Dec 27 '18 at 2:07









                        Nicolas FRANCOIS

                        3,6221516




                        3,6221516












                        • Um... you can't. But you can prove that the limits are all the same.
                          – fleablood
                          Dec 27 '18 at 2:26










                        • Yes you can : knowing that $(a_{2n})$ and $(a_{2n+1})$ both are convergent isn't enough (think of $(-1)^n$). But if you know that $(a_{3n})$ is also convergent, then you can prove that $(a_{2n})$ and $(a_{2n+1})$ have the same limit. Now THAT is enough to prove that $(a_n)$ is convergent.
                          – Nicolas FRANCOIS
                          Dec 28 '18 at 15:29










                        • Oh, I'm not sure how I misread you answer but I thought you hadn't including the $a_{2n+1}$ but just had the $a_{2n}$ and $a_{3n}$ as was the OP's question. As there are infinitely many terms belonging to both sequences they most converge to same. But, without, $a_{2n+1}$ there are infinitely many in neither so one couldn't conclude $a_n$ converge.
                          – fleablood
                          Dec 28 '18 at 16:47


















                        • Um... you can't. But you can prove that the limits are all the same.
                          – fleablood
                          Dec 27 '18 at 2:26










                        • Yes you can : knowing that $(a_{2n})$ and $(a_{2n+1})$ both are convergent isn't enough (think of $(-1)^n$). But if you know that $(a_{3n})$ is also convergent, then you can prove that $(a_{2n})$ and $(a_{2n+1})$ have the same limit. Now THAT is enough to prove that $(a_n)$ is convergent.
                          – Nicolas FRANCOIS
                          Dec 28 '18 at 15:29










                        • Oh, I'm not sure how I misread you answer but I thought you hadn't including the $a_{2n+1}$ but just had the $a_{2n}$ and $a_{3n}$ as was the OP's question. As there are infinitely many terms belonging to both sequences they most converge to same. But, without, $a_{2n+1}$ there are infinitely many in neither so one couldn't conclude $a_n$ converge.
                          – fleablood
                          Dec 28 '18 at 16:47
















                        Um... you can't. But you can prove that the limits are all the same.
                        – fleablood
                        Dec 27 '18 at 2:26




                        Um... you can't. But you can prove that the limits are all the same.
                        – fleablood
                        Dec 27 '18 at 2:26












                        Yes you can : knowing that $(a_{2n})$ and $(a_{2n+1})$ both are convergent isn't enough (think of $(-1)^n$). But if you know that $(a_{3n})$ is also convergent, then you can prove that $(a_{2n})$ and $(a_{2n+1})$ have the same limit. Now THAT is enough to prove that $(a_n)$ is convergent.
                        – Nicolas FRANCOIS
                        Dec 28 '18 at 15:29




                        Yes you can : knowing that $(a_{2n})$ and $(a_{2n+1})$ both are convergent isn't enough (think of $(-1)^n$). But if you know that $(a_{3n})$ is also convergent, then you can prove that $(a_{2n})$ and $(a_{2n+1})$ have the same limit. Now THAT is enough to prove that $(a_n)$ is convergent.
                        – Nicolas FRANCOIS
                        Dec 28 '18 at 15:29












                        Oh, I'm not sure how I misread you answer but I thought you hadn't including the $a_{2n+1}$ but just had the $a_{2n}$ and $a_{3n}$ as was the OP's question. As there are infinitely many terms belonging to both sequences they most converge to same. But, without, $a_{2n+1}$ there are infinitely many in neither so one couldn't conclude $a_n$ converge.
                        – fleablood
                        Dec 28 '18 at 16:47




                        Oh, I'm not sure how I misread you answer but I thought you hadn't including the $a_{2n+1}$ but just had the $a_{2n}$ and $a_{3n}$ as was the OP's question. As there are infinitely many terms belonging to both sequences they most converge to same. But, without, $a_{2n+1}$ there are infinitely many in neither so one couldn't conclude $a_n$ converge.
                        – fleablood
                        Dec 28 '18 at 16:47











                        0














                        The idea behind the proof of $ a_2n $ and $ a_(2n+1) $ being sufficient to conclude about the limit of $ a_n $ is that sequences of indices $ 2n $ and $ (2n+1) $ partition the family of indices $ I_n $, thus complete "the puzzle" if you wish. ie. they generate the family of indices $ I_n $, a subgroup of $ mathbb{Z}^+ $.



                        So, the procedure is without alternative; and the question about limit can be reduced to prove that **$<2n+1,3n>$ is also a generator for $ mathbb{Z}^+ $. To generate means that, there exists $ k_1,k_2 in mathbb{Z} $ such that:



                        $$ (1) iff k_1 (2n+1) + k_2 (3n) = N $$



                        where $ N $ is just any integer possible.



                        Observe that,



                        $$(1) iff n (2 k_1) + k_1 + n (3 k_2) = N $$



                        $$(1) iff n (2k_1 + 3k_2) + k_1 = N $$



                        We introduce: $ k_3 = 2k_1 + 3k_2 $:



                        $$(1) iff n k_3 + k_1 = N $$



                        In fact, we conclude that, for any $ k_3 $: $ <2n+1,3n> $ generates the quotient group $ mathbb{Z}/mathbb{Z}_(k_3) $. And, as we have no criteria over $k_3$, these two subsequences above are not sufficient to conclude anything about the main sequence's limit.






                        share|cite|improve this answer


























                          0














                          The idea behind the proof of $ a_2n $ and $ a_(2n+1) $ being sufficient to conclude about the limit of $ a_n $ is that sequences of indices $ 2n $ and $ (2n+1) $ partition the family of indices $ I_n $, thus complete "the puzzle" if you wish. ie. they generate the family of indices $ I_n $, a subgroup of $ mathbb{Z}^+ $.



                          So, the procedure is without alternative; and the question about limit can be reduced to prove that **$<2n+1,3n>$ is also a generator for $ mathbb{Z}^+ $. To generate means that, there exists $ k_1,k_2 in mathbb{Z} $ such that:



                          $$ (1) iff k_1 (2n+1) + k_2 (3n) = N $$



                          where $ N $ is just any integer possible.



                          Observe that,



                          $$(1) iff n (2 k_1) + k_1 + n (3 k_2) = N $$



                          $$(1) iff n (2k_1 + 3k_2) + k_1 = N $$



                          We introduce: $ k_3 = 2k_1 + 3k_2 $:



                          $$(1) iff n k_3 + k_1 = N $$



                          In fact, we conclude that, for any $ k_3 $: $ <2n+1,3n> $ generates the quotient group $ mathbb{Z}/mathbb{Z}_(k_3) $. And, as we have no criteria over $k_3$, these two subsequences above are not sufficient to conclude anything about the main sequence's limit.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            The idea behind the proof of $ a_2n $ and $ a_(2n+1) $ being sufficient to conclude about the limit of $ a_n $ is that sequences of indices $ 2n $ and $ (2n+1) $ partition the family of indices $ I_n $, thus complete "the puzzle" if you wish. ie. they generate the family of indices $ I_n $, a subgroup of $ mathbb{Z}^+ $.



                            So, the procedure is without alternative; and the question about limit can be reduced to prove that **$<2n+1,3n>$ is also a generator for $ mathbb{Z}^+ $. To generate means that, there exists $ k_1,k_2 in mathbb{Z} $ such that:



                            $$ (1) iff k_1 (2n+1) + k_2 (3n) = N $$



                            where $ N $ is just any integer possible.



                            Observe that,



                            $$(1) iff n (2 k_1) + k_1 + n (3 k_2) = N $$



                            $$(1) iff n (2k_1 + 3k_2) + k_1 = N $$



                            We introduce: $ k_3 = 2k_1 + 3k_2 $:



                            $$(1) iff n k_3 + k_1 = N $$



                            In fact, we conclude that, for any $ k_3 $: $ <2n+1,3n> $ generates the quotient group $ mathbb{Z}/mathbb{Z}_(k_3) $. And, as we have no criteria over $k_3$, these two subsequences above are not sufficient to conclude anything about the main sequence's limit.






                            share|cite|improve this answer












                            The idea behind the proof of $ a_2n $ and $ a_(2n+1) $ being sufficient to conclude about the limit of $ a_n $ is that sequences of indices $ 2n $ and $ (2n+1) $ partition the family of indices $ I_n $, thus complete "the puzzle" if you wish. ie. they generate the family of indices $ I_n $, a subgroup of $ mathbb{Z}^+ $.



                            So, the procedure is without alternative; and the question about limit can be reduced to prove that **$<2n+1,3n>$ is also a generator for $ mathbb{Z}^+ $. To generate means that, there exists $ k_1,k_2 in mathbb{Z} $ such that:



                            $$ (1) iff k_1 (2n+1) + k_2 (3n) = N $$



                            where $ N $ is just any integer possible.



                            Observe that,



                            $$(1) iff n (2 k_1) + k_1 + n (3 k_2) = N $$



                            $$(1) iff n (2k_1 + 3k_2) + k_1 = N $$



                            We introduce: $ k_3 = 2k_1 + 3k_2 $:



                            $$(1) iff n k_3 + k_1 = N $$



                            In fact, we conclude that, for any $ k_3 $: $ <2n+1,3n> $ generates the quotient group $ mathbb{Z}/mathbb{Z}_(k_3) $. And, as we have no criteria over $k_3$, these two subsequences above are not sufficient to conclude anything about the main sequence's limit.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 27 '18 at 16:28









                            freehumorist

                            121111




                            121111






















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