a sequence $frac19,frac{7}{17},frac{17}{25}, cdots$











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I have got a sequence $frac19,frac{7}{17},frac{17}{25}, cdots$ and got two different version for general term:



$$frac{2n^2-1}{8n+1} & frac{8(n-1)+(-1)^{n+1}}{8n+1}$$



but uisng the 1st one we get the sequence is divergent and the second one gives us convergence!!



I am really confused!










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  • the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
    – user8795
    Nov 17 at 3:09






  • 1




    Yes. You can show that fourth terms do not match so these are possibly two different sequences.
    – Anurag A
    Nov 17 at 3:12















up vote
-1
down vote

favorite












I have got a sequence $frac19,frac{7}{17},frac{17}{25}, cdots$ and got two different version for general term:



$$frac{2n^2-1}{8n+1} & frac{8(n-1)+(-1)^{n+1}}{8n+1}$$



but uisng the 1st one we get the sequence is divergent and the second one gives us convergence!!



I am really confused!










share|cite|improve this question






















  • the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
    – user8795
    Nov 17 at 3:09






  • 1




    Yes. You can show that fourth terms do not match so these are possibly two different sequences.
    – Anurag A
    Nov 17 at 3:12













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











I have got a sequence $frac19,frac{7}{17},frac{17}{25}, cdots$ and got two different version for general term:



$$frac{2n^2-1}{8n+1} & frac{8(n-1)+(-1)^{n+1}}{8n+1}$$



but uisng the 1st one we get the sequence is divergent and the second one gives us convergence!!



I am really confused!










share|cite|improve this question













I have got a sequence $frac19,frac{7}{17},frac{17}{25}, cdots$ and got two different version for general term:



$$frac{2n^2-1}{8n+1} & frac{8(n-1)+(-1)^{n+1}}{8n+1}$$



but uisng the 1st one we get the sequence is divergent and the second one gives us convergence!!



I am really confused!







real-analysis sequences-and-series






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asked Nov 17 at 3:01









user8795

5,57961846




5,57961846












  • the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
    – user8795
    Nov 17 at 3:09






  • 1




    Yes. You can show that fourth terms do not match so these are possibly two different sequences.
    – Anurag A
    Nov 17 at 3:12


















  • the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
    – user8795
    Nov 17 at 3:09






  • 1




    Yes. You can show that fourth terms do not match so these are possibly two different sequences.
    – Anurag A
    Nov 17 at 3:12
















the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
– user8795
Nov 17 at 3:09




the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
– user8795
Nov 17 at 3:09




1




1




Yes. You can show that fourth terms do not match so these are possibly two different sequences.
– Anurag A
Nov 17 at 3:12




Yes. You can show that fourth terms do not match so these are possibly two different sequences.
– Anurag A
Nov 17 at 3:12










3 Answers
3






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up vote
2
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These two expressions only agree until $n=3$. At $n=4$, the first expression gives $frac{31}{33}$, whereas the second gives $frac{23}{33}$. As $n to infty$, these expressions are giving you two different sequences. So one will converge and the other won't but their behaviors don't have to be dependent on each other just because the first few terms are a match.






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  • 1




    When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
    – Ross Millikan
    Nov 17 at 3:17




















up vote
0
down vote













This is possible, finitely many terms do not tell us about the convergence of the whole sequence.



Also, in general, we can't tell the whole sequence from finitely many terms.






share|cite|improve this answer




























    up vote
    0
    down vote













    You have made up your formulas based on just three terms. Others may come up with other formulas or just make up other terms for the rest of the sequence.



    You only have three terms of your sequence, depending on what happens with other terms everything is possible.



    For example you can define the rest of your sequence as $$4,5,6,...n,...$$ to make it divergent or $$ 0,0,0,...$$ or anything else.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

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      active

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      active

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      up vote
      2
      down vote













      These two expressions only agree until $n=3$. At $n=4$, the first expression gives $frac{31}{33}$, whereas the second gives $frac{23}{33}$. As $n to infty$, these expressions are giving you two different sequences. So one will converge and the other won't but their behaviors don't have to be dependent on each other just because the first few terms are a match.






      share|cite|improve this answer

















      • 1




        When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
        – Ross Millikan
        Nov 17 at 3:17

















      up vote
      2
      down vote













      These two expressions only agree until $n=3$. At $n=4$, the first expression gives $frac{31}{33}$, whereas the second gives $frac{23}{33}$. As $n to infty$, these expressions are giving you two different sequences. So one will converge and the other won't but their behaviors don't have to be dependent on each other just because the first few terms are a match.






      share|cite|improve this answer

















      • 1




        When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
        – Ross Millikan
        Nov 17 at 3:17















      up vote
      2
      down vote










      up vote
      2
      down vote









      These two expressions only agree until $n=3$. At $n=4$, the first expression gives $frac{31}{33}$, whereas the second gives $frac{23}{33}$. As $n to infty$, these expressions are giving you two different sequences. So one will converge and the other won't but their behaviors don't have to be dependent on each other just because the first few terms are a match.






      share|cite|improve this answer












      These two expressions only agree until $n=3$. At $n=4$, the first expression gives $frac{31}{33}$, whereas the second gives $frac{23}{33}$. As $n to infty$, these expressions are giving you two different sequences. So one will converge and the other won't but their behaviors don't have to be dependent on each other just because the first few terms are a match.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 17 at 3:06









      Anurag A

      24.9k12249




      24.9k12249








      • 1




        When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
        – Ross Millikan
        Nov 17 at 3:17
















      • 1




        When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
        – Ross Millikan
        Nov 17 at 3:17










      1




      1




      When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
      – Ross Millikan
      Nov 17 at 3:17






      When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
      – Ross Millikan
      Nov 17 at 3:17












      up vote
      0
      down vote













      This is possible, finitely many terms do not tell us about the convergence of the whole sequence.



      Also, in general, we can't tell the whole sequence from finitely many terms.






      share|cite|improve this answer

























        up vote
        0
        down vote













        This is possible, finitely many terms do not tell us about the convergence of the whole sequence.



        Also, in general, we can't tell the whole sequence from finitely many terms.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          This is possible, finitely many terms do not tell us about the convergence of the whole sequence.



          Also, in general, we can't tell the whole sequence from finitely many terms.






          share|cite|improve this answer












          This is possible, finitely many terms do not tell us about the convergence of the whole sequence.



          Also, in general, we can't tell the whole sequence from finitely many terms.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 17 at 3:03









          Siong Thye Goh

          94.6k1462114




          94.6k1462114






















              up vote
              0
              down vote













              You have made up your formulas based on just three terms. Others may come up with other formulas or just make up other terms for the rest of the sequence.



              You only have three terms of your sequence, depending on what happens with other terms everything is possible.



              For example you can define the rest of your sequence as $$4,5,6,...n,...$$ to make it divergent or $$ 0,0,0,...$$ or anything else.






              share|cite|improve this answer

























                up vote
                0
                down vote













                You have made up your formulas based on just three terms. Others may come up with other formulas or just make up other terms for the rest of the sequence.



                You only have three terms of your sequence, depending on what happens with other terms everything is possible.



                For example you can define the rest of your sequence as $$4,5,6,...n,...$$ to make it divergent or $$ 0,0,0,...$$ or anything else.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  You have made up your formulas based on just three terms. Others may come up with other formulas or just make up other terms for the rest of the sequence.



                  You only have three terms of your sequence, depending on what happens with other terms everything is possible.



                  For example you can define the rest of your sequence as $$4,5,6,...n,...$$ to make it divergent or $$ 0,0,0,...$$ or anything else.






                  share|cite|improve this answer












                  You have made up your formulas based on just three terms. Others may come up with other formulas or just make up other terms for the rest of the sequence.



                  You only have three terms of your sequence, depending on what happens with other terms everything is possible.



                  For example you can define the rest of your sequence as $$4,5,6,...n,...$$ to make it divergent or $$ 0,0,0,...$$ or anything else.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 17 at 3:13









                  Mohammad Riazi-Kermani

                  40.3k41958




                  40.3k41958






























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