Can $a(n) = frac{n}{n+1}$ be written recursively?












1












$begingroup$


Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$



Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$



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







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    1












    $begingroup$


    Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$



    Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$



    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.







    $endgroup$















      1












      1








      1





      $begingroup$


      Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$



      Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$



      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.







      $endgroup$




      Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$



      Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$



      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




      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.
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      share|cite|improve this question








      edited 9 mins ago









      user1952500

      831712




      831712






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      asked 1 hour ago









      Levi KLevi K

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          3 Answers
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          $begingroup$

          begin{align*}
          a_{n+1} &= frac{n+1}{n+2} \
          &= frac{n+2-1}{n+2} \
          &= 1 - frac{1}{n+2} text{, so } \
          1 - a_{n+1} &= frac{1}{n+2} text{, } \
          frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
          &= n+1+1 \
          &= frac{1}{1- a_n} +1 \
          &= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
          &= frac{2-a_n}{1- a_n} text{, then } \
          1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
          a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
          &= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
          &= frac{1}{2- a_n} text{.}
          end{align*}






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            2












            $begingroup$

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



            It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$






            share|cite|improve this answer










            New contributor




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





















              0












              $begingroup$

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



              $$a_n = frac{na_{n-1} + 1}{n+1}$$



              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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                active

                oldest

                votes








                3 Answers
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                2












                $begingroup$

                begin{align*}
                a_{n+1} &= frac{n+1}{n+2} \
                &= frac{n+2-1}{n+2} \
                &= 1 - frac{1}{n+2} text{, so } \
                1 - a_{n+1} &= frac{1}{n+2} text{, } \
                frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
                &= n+1+1 \
                &= frac{1}{1- a_n} +1 \
                &= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
                &= frac{2-a_n}{1- a_n} text{, then } \
                1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
                a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
                &= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
                &= frac{1}{2- a_n} text{.}
                end{align*}






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                $endgroup$


















                  2












                  $begingroup$

                  begin{align*}
                  a_{n+1} &= frac{n+1}{n+2} \
                  &= frac{n+2-1}{n+2} \
                  &= 1 - frac{1}{n+2} text{, so } \
                  1 - a_{n+1} &= frac{1}{n+2} text{, } \
                  frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
                  &= n+1+1 \
                  &= frac{1}{1- a_n} +1 \
                  &= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
                  &= frac{2-a_n}{1- a_n} text{, then } \
                  1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
                  a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
                  &= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
                  &= frac{1}{2- a_n} text{.}
                  end{align*}






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    begin{align*}
                    a_{n+1} &= frac{n+1}{n+2} \
                    &= frac{n+2-1}{n+2} \
                    &= 1 - frac{1}{n+2} text{, so } \
                    1 - a_{n+1} &= frac{1}{n+2} text{, } \
                    frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
                    &= n+1+1 \
                    &= frac{1}{1- a_n} +1 \
                    &= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
                    &= frac{2-a_n}{1- a_n} text{, then } \
                    1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
                    a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
                    &= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
                    &= frac{1}{2- a_n} text{.}
                    end{align*}






                    share|cite|improve this answer









                    $endgroup$



                    begin{align*}
                    a_{n+1} &= frac{n+1}{n+2} \
                    &= frac{n+2-1}{n+2} \
                    &= 1 - frac{1}{n+2} text{, so } \
                    1 - a_{n+1} &= frac{1}{n+2} text{, } \
                    frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
                    &= n+1+1 \
                    &= frac{1}{1- a_n} +1 \
                    &= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
                    &= frac{2-a_n}{1- a_n} text{, then } \
                    1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
                    a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
                    &= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
                    &= frac{1}{2- a_n} text{.}
                    end{align*}







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                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 43 mins ago









                    Eric TowersEric Towers

                    33.5k22370




                    33.5k22370























                        2












                        $begingroup$

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



                        It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$






                        share|cite|improve this answer










                        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












                          $begingroup$

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



                          It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$






                          share|cite|improve this answer










                          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 $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$






                            share|cite|improve this answer










                            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 $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$







                            share|cite|improve this answer










                            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 answer



                            share|cite|improve this answer








                            edited 11 secs ago









                            user1952500

                            831712




                            831712






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                            answered 1 hour 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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                                0












                                $begingroup$

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



                                $$a_n = frac{na_{n-1} + 1}{n+1}$$



                                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



                                  $$a_n = frac{na_{n-1} + 1}{n+1}$$



                                  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



                                    $$a_n = frac{na_{n-1} + 1}{n+1}$$



                                    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



                                    $$a_n = frac{na_{n-1} + 1}{n+1}$$



                                    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

                                    9,91731740




                                    9,91731740






















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