Finding the monotonicity of simple sequence - how to?












4












$begingroup$


I'm trying to find the monotonicity (whether it's increasing, decreasing or non-existeng) of such simple sequence: $$a_{n} = sqrt[n]{2^n+3^n}$$



$$frac{a_{n}}{a_{n+1}}=frac{sqrt[n]{2^n+3^n}}{sqrt[n+1]{2^{n+1}+3^{n+1}}}$$



I have dealt with such exercises before without problems. This one I have no idea how to proceed further.



Help is appreciated, thanks.










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  • $begingroup$
    Easily seen from $a_n=3sqrt[n]{(2/3)^n+1}$.
    $endgroup$
    – metamorphy
    Jan 2 at 0:06
















4












$begingroup$


I'm trying to find the monotonicity (whether it's increasing, decreasing or non-existeng) of such simple sequence: $$a_{n} = sqrt[n]{2^n+3^n}$$



$$frac{a_{n}}{a_{n+1}}=frac{sqrt[n]{2^n+3^n}}{sqrt[n+1]{2^{n+1}+3^{n+1}}}$$



I have dealt with such exercises before without problems. This one I have no idea how to proceed further.



Help is appreciated, thanks.










share|cite|improve this question









$endgroup$












  • $begingroup$
    Easily seen from $a_n=3sqrt[n]{(2/3)^n+1}$.
    $endgroup$
    – metamorphy
    Jan 2 at 0:06














4












4








4





$begingroup$


I'm trying to find the monotonicity (whether it's increasing, decreasing or non-existeng) of such simple sequence: $$a_{n} = sqrt[n]{2^n+3^n}$$



$$frac{a_{n}}{a_{n+1}}=frac{sqrt[n]{2^n+3^n}}{sqrt[n+1]{2^{n+1}+3^{n+1}}}$$



I have dealt with such exercises before without problems. This one I have no idea how to proceed further.



Help is appreciated, thanks.










share|cite|improve this question









$endgroup$




I'm trying to find the monotonicity (whether it's increasing, decreasing or non-existeng) of such simple sequence: $$a_{n} = sqrt[n]{2^n+3^n}$$



$$frac{a_{n}}{a_{n+1}}=frac{sqrt[n]{2^n+3^n}}{sqrt[n+1]{2^{n+1}+3^{n+1}}}$$



I have dealt with such exercises before without problems. This one I have no idea how to proceed further.



Help is appreciated, thanks.







real-analysis sequences-and-series limits






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asked Jan 1 at 23:42









JanPawelJanPawel

211




211












  • $begingroup$
    Easily seen from $a_n=3sqrt[n]{(2/3)^n+1}$.
    $endgroup$
    – metamorphy
    Jan 2 at 0:06


















  • $begingroup$
    Easily seen from $a_n=3sqrt[n]{(2/3)^n+1}$.
    $endgroup$
    – metamorphy
    Jan 2 at 0:06
















$begingroup$
Easily seen from $a_n=3sqrt[n]{(2/3)^n+1}$.
$endgroup$
– metamorphy
Jan 2 at 0:06




$begingroup$
Easily seen from $a_n=3sqrt[n]{(2/3)^n+1}$.
$endgroup$
– metamorphy
Jan 2 at 0:06










1 Answer
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$begingroup$

Rearrange things a bit



$$
a_n = 3left[left(frac{2}{3}right)^n + 1 right]^{1/n}
$$



As $n$ increases the term $(2/3)^n$ goes to zero and $a_n$ monotonically decreases to $3$






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Rearrange things a bit



    $$
    a_n = 3left[left(frac{2}{3}right)^n + 1 right]^{1/n}
    $$



    As $n$ increases the term $(2/3)^n$ goes to zero and $a_n$ monotonically decreases to $3$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Rearrange things a bit



      $$
      a_n = 3left[left(frac{2}{3}right)^n + 1 right]^{1/n}
      $$



      As $n$ increases the term $(2/3)^n$ goes to zero and $a_n$ monotonically decreases to $3$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Rearrange things a bit



        $$
        a_n = 3left[left(frac{2}{3}right)^n + 1 right]^{1/n}
        $$



        As $n$ increases the term $(2/3)^n$ goes to zero and $a_n$ monotonically decreases to $3$






        share|cite|improve this answer









        $endgroup$



        Rearrange things a bit



        $$
        a_n = 3left[left(frac{2}{3}right)^n + 1 right]^{1/n}
        $$



        As $n$ increases the term $(2/3)^n$ goes to zero and $a_n$ monotonically decreases to $3$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 0:08









        caveraccaverac

        14.5k31130




        14.5k31130






























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