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Can $a(n) = fracnn+1$ be written recursively?


Formula for a sequenceUse of Recursively Defined FunctionsRecursive Sequence from Finite SequencesFinding the lowest common value in repeating sequencesUnexplanied pattern from increasing rational sequencesWhat would describe the following basic sequence?Understanding sub-sequencesCan the Fibonacci sequence be written as an explicit rule?Turning a recursively defined sequence into an explicit formulaWhat's the formula for producing these series?













1












$begingroup$


Take the sequence $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$



Algebraically it can be written as $$a(n) = fracnn + 1$$



Can you write this as a recursive function as well?



A pattern I have noticed:



  • Take $A_n-1$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.

I am currently in Algebra II Honors and learning sequences










share|cite|improve this question









New contributor




Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
















    1












    $begingroup$


    Take the sequence $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$



    Algebraically it can be written as $$a(n) = fracnn + 1$$



    Can you write this as a recursive function as well?



    A pattern I have noticed:



    • Take $A_n-1$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.

    I am currently in Algebra II Honors and learning sequences










    share|cite|improve this question









    New contributor




    Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      1












      1








      1





      $begingroup$


      Take the sequence $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$



      Algebraically it can be written as $$a(n) = fracnn + 1$$



      Can you write this as a recursive function as well?



      A pattern I have noticed:



      • Take $A_n-1$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.

      I am currently in Algebra II Honors and learning sequences










      share|cite|improve this question









      New contributor




      Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Take the sequence $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$



      Algebraically it can be written as $$a(n) = fracnn + 1$$



      Can you write this as a recursive function as well?



      A pattern I have noticed:



      • Take $A_n-1$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.

      I am currently in Algebra II Honors and learning sequences







      sequences-and-series number-theory recursion






      share|cite|improve this question









      New contributor




      Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 6 mins ago









      user1952500

      829712




      829712






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      asked 1 hour ago









      Levi KLevi K

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          3 Answers
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          2












          $begingroup$

          After some further solving, I was able to come up with an answer



          It can be written $$A_n + 1 = (2 - A_n)^-1$$ where $$A_1 = 1/2$$






          share|cite








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          Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          $endgroup$




















            2












            $begingroup$

            beginalign*
            a_n+1 &= fracn+1n+2 \
            &= fracn+2-1n+2 \
            &= 1 - frac1n+2 text, so \
            1 - a_n+1 &= frac1n+2 text, \
            frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
            &= n+1+1 \
            &= frac11- a_n +1 \
            &= frac11- a_n + frac1-a_n1-a_n \
            &= frac2-a_n1- a_n text, then \
            1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
            a_n+1 &= 1 - frac1-a_n2- a_n \
            &= frac2-a_n2- a_n - frac1-a_n2- a_n \
            &= frac12- a_n text.
            endalign*






            share|cite|improve this answer









            $endgroup$




















              0












              $begingroup$

              Just by playing around with some numbers, I determined a recursive relation to be



              $$a_n = fracna_n-1 + 1n+1$$



              with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).






              share|cite|improve this answer









              $endgroup$













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                3 Answers
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                3 Answers
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                active

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                active

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                2












                $begingroup$

                After some further solving, I was able to come up with an answer



                It can be written $$A_n + 1 = (2 - A_n)^-1$$ where $$A_1 = 1/2$$






                share|cite








                New contributor




                Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                $endgroup$

















                  2












                  $begingroup$

                  After some further solving, I was able to come up with an answer



                  It can be written $$A_n + 1 = (2 - A_n)^-1$$ where $$A_1 = 1/2$$






                  share|cite








                  New contributor




                  Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    After some further solving, I was able to come up with an answer



                    It can be written $$A_n + 1 = (2 - A_n)^-1$$ where $$A_1 = 1/2$$






                    share|cite








                    New contributor




                    Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    After some further solving, I was able to come up with an answer



                    It can be written $$A_n + 1 = (2 - A_n)^-1$$ where $$A_1 = 1/2$$







                    share|cite








                    New contributor




                    Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    share|cite



                    share|cite






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                    answered 59 mins ago









                    Levi KLevi K

                    262




                    262




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                    New contributor





                    Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                        2












                        $begingroup$

                        beginalign*
                        a_n+1 &= fracn+1n+2 \
                        &= fracn+2-1n+2 \
                        &= 1 - frac1n+2 text, so \
                        1 - a_n+1 &= frac1n+2 text, \
                        frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
                        &= n+1+1 \
                        &= frac11- a_n +1 \
                        &= frac11- a_n + frac1-a_n1-a_n \
                        &= frac2-a_n1- a_n text, then \
                        1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
                        a_n+1 &= 1 - frac1-a_n2- a_n \
                        &= frac2-a_n2- a_n - frac1-a_n2- a_n \
                        &= frac12- a_n text.
                        endalign*






                        share|cite|improve this answer









                        $endgroup$

















                          2












                          $begingroup$

                          beginalign*
                          a_n+1 &= fracn+1n+2 \
                          &= fracn+2-1n+2 \
                          &= 1 - frac1n+2 text, so \
                          1 - a_n+1 &= frac1n+2 text, \
                          frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
                          &= n+1+1 \
                          &= frac11- a_n +1 \
                          &= frac11- a_n + frac1-a_n1-a_n \
                          &= frac2-a_n1- a_n text, then \
                          1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
                          a_n+1 &= 1 - frac1-a_n2- a_n \
                          &= frac2-a_n2- a_n - frac1-a_n2- a_n \
                          &= frac12- a_n text.
                          endalign*






                          share|cite|improve this answer









                          $endgroup$















                            2












                            2








                            2





                            $begingroup$

                            beginalign*
                            a_n+1 &= fracn+1n+2 \
                            &= fracn+2-1n+2 \
                            &= 1 - frac1n+2 text, so \
                            1 - a_n+1 &= frac1n+2 text, \
                            frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
                            &= n+1+1 \
                            &= frac11- a_n +1 \
                            &= frac11- a_n + frac1-a_n1-a_n \
                            &= frac2-a_n1- a_n text, then \
                            1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
                            a_n+1 &= 1 - frac1-a_n2- a_n \
                            &= frac2-a_n2- a_n - frac1-a_n2- a_n \
                            &= frac12- a_n text.
                            endalign*






                            share|cite|improve this answer









                            $endgroup$



                            beginalign*
                            a_n+1 &= fracn+1n+2 \
                            &= fracn+2-1n+2 \
                            &= 1 - frac1n+2 text, so \
                            1 - a_n+1 &= frac1n+2 text, \
                            frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
                            &= n+1+1 \
                            &= frac11- a_n +1 \
                            &= frac11- a_n + frac1-a_n1-a_n \
                            &= frac2-a_n1- a_n text, then \
                            1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
                            a_n+1 &= 1 - frac1-a_n2- a_n \
                            &= frac2-a_n2- a_n - frac1-a_n2- a_n \
                            &= frac12- a_n text.
                            endalign*







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 41 mins ago









                            Eric TowersEric Towers

                            33.5k22370




                            33.5k22370





















                                0












                                $begingroup$

                                Just by playing around with some numbers, I determined a recursive relation to be



                                $$a_n = fracna_n-1 + 1n+1$$



                                with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).






                                share|cite|improve this answer









                                $endgroup$

















                                  0












                                  $begingroup$

                                  Just by playing around with some numbers, I determined a recursive relation to be



                                  $$a_n = fracna_n-1 + 1n+1$$



                                  with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).






                                  share|cite|improve this answer









                                  $endgroup$















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Just by playing around with some numbers, I determined a recursive relation to be



                                    $$a_n = fracna_n-1 + 1n+1$$



                                    with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).






                                    share|cite|improve this answer









                                    $endgroup$



                                    Just by playing around with some numbers, I determined a recursive relation to be



                                    $$a_n = fracna_n-1 + 1n+1$$



                                    with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 1 hour ago









                                    Eevee TrainerEevee Trainer

                                    9,91631740




                                    9,91631740




















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