Find maximum of succession $a_n+1=fracn^2+n+42n^2+1a_n$ knowing $a_1=1$

Multi tool use
Multi tool use

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












I want to find out what's the maximum of the following sequence:



$$
left{
beginarrayc
a_1=1 \
a_n+1=fracn^2+n+42n^2+1a_n \
endarray
right.
$$



I know that the limit of $a_n$ is equal to $0$. I have no clue about how to find the maximum. I have tried to find out the values of $a_1, a_2, a_3dots$ but we couldn't find the maximum. Please note that $ngeq 0$. Any hints?










share|cite|improve this question



















  • 1




    There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
    – José Carlos Santos
    Sep 10 at 9:59










  • @JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
    – Cesare
    Sep 10 at 10:01







  • 1




    So, you are sure that the definition of $a_n+1$ does not use $a_n$.
    – José Carlos Santos
    Sep 10 at 10:02










  • It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
    – Claude Leibovici
    Sep 10 at 10:03










  • @JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
    – Cesare
    Sep 10 at 10:05














up vote
1
down vote

favorite












I want to find out what's the maximum of the following sequence:



$$
left{
beginarrayc
a_1=1 \
a_n+1=fracn^2+n+42n^2+1a_n \
endarray
right.
$$



I know that the limit of $a_n$ is equal to $0$. I have no clue about how to find the maximum. I have tried to find out the values of $a_1, a_2, a_3dots$ but we couldn't find the maximum. Please note that $ngeq 0$. Any hints?










share|cite|improve this question



















  • 1




    There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
    – José Carlos Santos
    Sep 10 at 9:59










  • @JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
    – Cesare
    Sep 10 at 10:01







  • 1




    So, you are sure that the definition of $a_n+1$ does not use $a_n$.
    – José Carlos Santos
    Sep 10 at 10:02










  • It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
    – Claude Leibovici
    Sep 10 at 10:03










  • @JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
    – Cesare
    Sep 10 at 10:05












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I want to find out what's the maximum of the following sequence:



$$
left{
beginarrayc
a_1=1 \
a_n+1=fracn^2+n+42n^2+1a_n \
endarray
right.
$$



I know that the limit of $a_n$ is equal to $0$. I have no clue about how to find the maximum. I have tried to find out the values of $a_1, a_2, a_3dots$ but we couldn't find the maximum. Please note that $ngeq 0$. Any hints?










share|cite|improve this question















I want to find out what's the maximum of the following sequence:



$$
left{
beginarrayc
a_1=1 \
a_n+1=fracn^2+n+42n^2+1a_n \
endarray
right.
$$



I know that the limit of $a_n$ is equal to $0$. I have no clue about how to find the maximum. I have tried to find out the values of $a_1, a_2, a_3dots$ but we couldn't find the maximum. Please note that $ngeq 0$. Any hints?







sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 10 at 11:29

























asked Sep 10 at 9:58









Cesare

734410




734410







  • 1




    There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
    – José Carlos Santos
    Sep 10 at 9:59










  • @JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
    – Cesare
    Sep 10 at 10:01







  • 1




    So, you are sure that the definition of $a_n+1$ does not use $a_n$.
    – José Carlos Santos
    Sep 10 at 10:02










  • It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
    – Claude Leibovici
    Sep 10 at 10:03










  • @JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
    – Cesare
    Sep 10 at 10:05












  • 1




    There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
    – José Carlos Santos
    Sep 10 at 9:59










  • @JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
    – Cesare
    Sep 10 at 10:01







  • 1




    So, you are sure that the definition of $a_n+1$ does not use $a_n$.
    – José Carlos Santos
    Sep 10 at 10:02










  • It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
    – Claude Leibovici
    Sep 10 at 10:03










  • @JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
    – Cesare
    Sep 10 at 10:05







1




1




There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
– José Carlos Santos
Sep 10 at 9:59




There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
– José Carlos Santos
Sep 10 at 9:59












@JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
– Cesare
Sep 10 at 10:01





@JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
– Cesare
Sep 10 at 10:01





1




1




So, you are sure that the definition of $a_n+1$ does not use $a_n$.
– José Carlos Santos
Sep 10 at 10:02




So, you are sure that the definition of $a_n+1$ does not use $a_n$.
– José Carlos Santos
Sep 10 at 10:02












It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
– Claude Leibovici
Sep 10 at 10:03




It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
– Claude Leibovici
Sep 10 at 10:03












@JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
– Cesare
Sep 10 at 10:05




@JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
– Cesare
Sep 10 at 10:05










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










It follows from your definition that $a_n+1>a_n$ is $n=1$ or $n=2$ and that $a_n+1<a_n$ otherwise. Therefore, the maximum is attained when $n=3$. That maximum is $frac209$, since $a_3=frac209$.






share|cite|improve this answer






















  • My textbook says that the extreme value of the function (maximum) is $frac 2019$.
    – Cesare
    Sep 10 at 10:16











  • @Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
    – José Carlos Santos
    Sep 10 at 10:19










  • @Winther I saw that. Why do you think otherwise?
    – José Carlos Santos
    Sep 10 at 10:20










  • @Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
    – José Carlos Santos
    Sep 10 at 10:25










  • Therefore $a_4$ should be less than $frac 209$?
    – Cesare
    Sep 10 at 10:27


















up vote
1
down vote













In general for a recursion of the form:
$$
a_n+1 = f(n)a_n, a_1 > 0.
$$
if $f(n) - 1$ has only one zero $> 0$ and $lim_ntoinftyf(n) < 1$ then you know $a_n$ will attain its maximum for the largest $n$ for which you have $f(n)geq 1$.



in this case:
$f(n)= fracn^2+n+42n^2+1 > 1Rightarrow - n^2 +n +3 > 0 Rightarrow n < frac1+sqrt132 approx 2.303$. Thus your maximum is $a_3$ ...






share|cite|improve this answer




















    Your Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2911750%2ffind-maximum-of-succession-a-n1-fracn2n42n21a-n-knowing-a-1-1%23new-answer', 'question_page');

    );

    Post as a guest






























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    It follows from your definition that $a_n+1>a_n$ is $n=1$ or $n=2$ and that $a_n+1<a_n$ otherwise. Therefore, the maximum is attained when $n=3$. That maximum is $frac209$, since $a_3=frac209$.






    share|cite|improve this answer






















    • My textbook says that the extreme value of the function (maximum) is $frac 2019$.
      – Cesare
      Sep 10 at 10:16











    • @Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
      – José Carlos Santos
      Sep 10 at 10:19










    • @Winther I saw that. Why do you think otherwise?
      – José Carlos Santos
      Sep 10 at 10:20










    • @Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
      – José Carlos Santos
      Sep 10 at 10:25










    • Therefore $a_4$ should be less than $frac 209$?
      – Cesare
      Sep 10 at 10:27















    up vote
    1
    down vote



    accepted










    It follows from your definition that $a_n+1>a_n$ is $n=1$ or $n=2$ and that $a_n+1<a_n$ otherwise. Therefore, the maximum is attained when $n=3$. That maximum is $frac209$, since $a_3=frac209$.






    share|cite|improve this answer






















    • My textbook says that the extreme value of the function (maximum) is $frac 2019$.
      – Cesare
      Sep 10 at 10:16











    • @Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
      – José Carlos Santos
      Sep 10 at 10:19










    • @Winther I saw that. Why do you think otherwise?
      – José Carlos Santos
      Sep 10 at 10:20










    • @Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
      – José Carlos Santos
      Sep 10 at 10:25










    • Therefore $a_4$ should be less than $frac 209$?
      – Cesare
      Sep 10 at 10:27













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    It follows from your definition that $a_n+1>a_n$ is $n=1$ or $n=2$ and that $a_n+1<a_n$ otherwise. Therefore, the maximum is attained when $n=3$. That maximum is $frac209$, since $a_3=frac209$.






    share|cite|improve this answer














    It follows from your definition that $a_n+1>a_n$ is $n=1$ or $n=2$ and that $a_n+1<a_n$ otherwise. Therefore, the maximum is attained when $n=3$. That maximum is $frac209$, since $a_3=frac209$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 10 at 10:24

























    answered Sep 10 at 10:13









    José Carlos Santos

    124k17101186




    124k17101186











    • My textbook says that the extreme value of the function (maximum) is $frac 2019$.
      – Cesare
      Sep 10 at 10:16











    • @Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
      – José Carlos Santos
      Sep 10 at 10:19










    • @Winther I saw that. Why do you think otherwise?
      – José Carlos Santos
      Sep 10 at 10:20










    • @Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
      – José Carlos Santos
      Sep 10 at 10:25










    • Therefore $a_4$ should be less than $frac 209$?
      – Cesare
      Sep 10 at 10:27

















    • My textbook says that the extreme value of the function (maximum) is $frac 2019$.
      – Cesare
      Sep 10 at 10:16











    • @Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
      – José Carlos Santos
      Sep 10 at 10:19










    • @Winther I saw that. Why do you think otherwise?
      – José Carlos Santos
      Sep 10 at 10:20










    • @Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
      – José Carlos Santos
      Sep 10 at 10:25










    • Therefore $a_4$ should be less than $frac 209$?
      – Cesare
      Sep 10 at 10:27
















    My textbook says that the extreme value of the function (maximum) is $frac 2019$.
    – Cesare
    Sep 10 at 10:16





    My textbook says that the extreme value of the function (maximum) is $frac 2019$.
    – Cesare
    Sep 10 at 10:16













    @Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
    – José Carlos Santos
    Sep 10 at 10:19




    @Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
    – José Carlos Santos
    Sep 10 at 10:19












    @Winther I saw that. Why do you think otherwise?
    – José Carlos Santos
    Sep 10 at 10:20




    @Winther I saw that. Why do you think otherwise?
    – José Carlos Santos
    Sep 10 at 10:20












    @Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
    – José Carlos Santos
    Sep 10 at 10:25




    @Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
    – José Carlos Santos
    Sep 10 at 10:25












    Therefore $a_4$ should be less than $frac 209$?
    – Cesare
    Sep 10 at 10:27





    Therefore $a_4$ should be less than $frac 209$?
    – Cesare
    Sep 10 at 10:27











    up vote
    1
    down vote













    In general for a recursion of the form:
    $$
    a_n+1 = f(n)a_n, a_1 > 0.
    $$
    if $f(n) - 1$ has only one zero $> 0$ and $lim_ntoinftyf(n) < 1$ then you know $a_n$ will attain its maximum for the largest $n$ for which you have $f(n)geq 1$.



    in this case:
    $f(n)= fracn^2+n+42n^2+1 > 1Rightarrow - n^2 +n +3 > 0 Rightarrow n < frac1+sqrt132 approx 2.303$. Thus your maximum is $a_3$ ...






    share|cite|improve this answer
























      up vote
      1
      down vote













      In general for a recursion of the form:
      $$
      a_n+1 = f(n)a_n, a_1 > 0.
      $$
      if $f(n) - 1$ has only one zero $> 0$ and $lim_ntoinftyf(n) < 1$ then you know $a_n$ will attain its maximum for the largest $n$ for which you have $f(n)geq 1$.



      in this case:
      $f(n)= fracn^2+n+42n^2+1 > 1Rightarrow - n^2 +n +3 > 0 Rightarrow n < frac1+sqrt132 approx 2.303$. Thus your maximum is $a_3$ ...






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        In general for a recursion of the form:
        $$
        a_n+1 = f(n)a_n, a_1 > 0.
        $$
        if $f(n) - 1$ has only one zero $> 0$ and $lim_ntoinftyf(n) < 1$ then you know $a_n$ will attain its maximum for the largest $n$ for which you have $f(n)geq 1$.



        in this case:
        $f(n)= fracn^2+n+42n^2+1 > 1Rightarrow - n^2 +n +3 > 0 Rightarrow n < frac1+sqrt132 approx 2.303$. Thus your maximum is $a_3$ ...






        share|cite|improve this answer












        In general for a recursion of the form:
        $$
        a_n+1 = f(n)a_n, a_1 > 0.
        $$
        if $f(n) - 1$ has only one zero $> 0$ and $lim_ntoinftyf(n) < 1$ then you know $a_n$ will attain its maximum for the largest $n$ for which you have $f(n)geq 1$.



        in this case:
        $f(n)= fracn^2+n+42n^2+1 > 1Rightarrow - n^2 +n +3 > 0 Rightarrow n < frac1+sqrt132 approx 2.303$. Thus your maximum is $a_3$ ...







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 10 at 10:25









        denklo

        3335




        3335



























             

            draft saved


            draft discarded















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2911750%2ffind-maximum-of-succession-a-n1-fracn2n42n21a-n-knowing-a-1-1%23new-answer', 'question_page');

            );

            Post as a guest













































































            Uc,hQIna90vd9hnAGJ9 E9kp,rzOnsSi Rp7,PqENjS,wwSZu0,tUREsY2 MPC O7crkh7CZ7 FS h,yfBd,YMIhGU07mpXA6KqQ6oBK
            1JJj yoKq9hgktsiLqrra5IgI,JopYAr61XOc,xaizFTv,q5r5S

            這個網誌中的熱門文章

            How to combine Bézier curves to a surface?

            Propositional logic and tautologies

            Distribution of Stopped Wiener Process with Stochastic Volatility