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.
      Check out our Code of Conduct.









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      edited 7 mins ago









      user1952500

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      asked 1 hour ago









      Levi KLevi K

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          $begingroup$

          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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            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









            $endgroup$





















              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

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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.
                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 $${A_{n + 1}} = ({2 - A_{n}})^{-1}$$ 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 1 hour ago









                    Levi KLevi K

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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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                            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



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



                                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

                                    9,91631740




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