What is the maximum value of $frac{x^{100}}{1+x+x^2+ldots+x^{200}}$?












7












$begingroup$


If $x$ is positive, what is the maximum value of this expression:



$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$



This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.



This is what I have tried:



The denominator is a geometric series whose sum is



$$frac{1-x^{201}}{1-x}$$



The fraction now becomes



$$frac{x^{100}(1-x)}{1-x^{201}}$$



I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.



That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).



I can't see such means from looking at the fraction. Can someone help me here?






inequality means geometric-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
    $endgroup$
    – Mark Viola
    Jan 30 at 4:45


















7












$begingroup$


If $x$ is positive, what is the maximum value of this expression:



$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$



This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.



This is what I have tried:



The denominator is a geometric series whose sum is



$$frac{1-x^{201}}{1-x}$$



The fraction now becomes



$$frac{x^{100}(1-x)}{1-x^{201}}$$



I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.



That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).



I can't see such means from looking at the fraction. Can someone help me here?






inequality means geometric-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
    $endgroup$
    – Mark Viola
    Jan 30 at 4:45
















7












7








7


2



$begingroup$


If $x$ is positive, what is the maximum value of this expression:



$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$



This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.



This is what I have tried:



The denominator is a geometric series whose sum is



$$frac{1-x^{201}}{1-x}$$



The fraction now becomes



$$frac{x^{100}(1-x)}{1-x^{201}}$$



I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.



That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).



I can't see such means from looking at the fraction. Can someone help me here?






inequality means geometric-series







share|cite|improve this question











$endgroup$




If $x$ is positive, what is the maximum value of this expression:



$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$



This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.



This is what I have tried:



The denominator is a geometric series whose sum is



$$frac{1-x^{201}}{1-x}$$



The fraction now becomes



$$frac{x^{100}(1-x)}{1-x^{201}}$$



I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.



That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).



I can't see such means from looking at the fraction. Can someone help me here?






inequality means geometric-series




inequality means geometric-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 11:39









user21820

40k544160




40k544160










asked Jan 30 at 4:25









user69284user69284

17918




17918








  • 2




    $begingroup$
    For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
    $endgroup$
    – Mark Viola
    Jan 30 at 4:45
















  • 2




    $begingroup$
    For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
    $endgroup$
    – Mark Viola
    Jan 30 at 4:45










2




2




$begingroup$
For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
– Mark Viola
Jan 30 at 4:45






$begingroup$
For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
– Mark Viola
Jan 30 at 4:45












3 Answers
3






active

oldest

votes


















6












$begingroup$

The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.






share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    You can instead minimise the reciprocal of your quantity, viz.,
    $$frac{1+x+x^2+cdots+x^{200}}{x^{100}}=x^{-100}+x^{-99}+cdots+x^{99}+x^{100}.$$
    One only needs the two-variable AM/GM inequality to do this, just in the
    form $y+y^{-1}ge2$ for $y>0$, for
    $$x^{-100}+x^{-99}+cdots+x^{99}+x^{100}=1+sum_{n=1}^{100}(x^n+x^{-n})
    ge201$$
    with equality if $x=1$.






    share|cite|improve this answer









    $endgroup$





















      4












      $begingroup$

      For $x>0$, we have from the AM-GM inequality



      $$begin{align}
      sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
      &=201 sqrt[201]{x^{20100}}\\
      &=201x^{100}
      end{align}$$



      Hence, we see that



      $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Please let me know how I can improve my answer. I really want to give you the best answer I can.
        $endgroup$
        – Mark Viola
        Jan 30 at 5:02












      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.






      share|cite|improve this answer









      $endgroup$


















        6












        $begingroup$

        The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.






        share|cite|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.






          share|cite|improve this answer









          $endgroup$



          The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 30 at 4:42









          jmerryjmerry

          17k11633




          17k11633























              5












              $begingroup$

              You can instead minimise the reciprocal of your quantity, viz.,
              $$frac{1+x+x^2+cdots+x^{200}}{x^{100}}=x^{-100}+x^{-99}+cdots+x^{99}+x^{100}.$$
              One only needs the two-variable AM/GM inequality to do this, just in the
              form $y+y^{-1}ge2$ for $y>0$, for
              $$x^{-100}+x^{-99}+cdots+x^{99}+x^{100}=1+sum_{n=1}^{100}(x^n+x^{-n})
              ge201$$
              with equality if $x=1$.






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                You can instead minimise the reciprocal of your quantity, viz.,
                $$frac{1+x+x^2+cdots+x^{200}}{x^{100}}=x^{-100}+x^{-99}+cdots+x^{99}+x^{100}.$$
                One only needs the two-variable AM/GM inequality to do this, just in the
                form $y+y^{-1}ge2$ for $y>0$, for
                $$x^{-100}+x^{-99}+cdots+x^{99}+x^{100}=1+sum_{n=1}^{100}(x^n+x^{-n})
                ge201$$
                with equality if $x=1$.






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  You can instead minimise the reciprocal of your quantity, viz.,
                  $$frac{1+x+x^2+cdots+x^{200}}{x^{100}}=x^{-100}+x^{-99}+cdots+x^{99}+x^{100}.$$
                  One only needs the two-variable AM/GM inequality to do this, just in the
                  form $y+y^{-1}ge2$ for $y>0$, for
                  $$x^{-100}+x^{-99}+cdots+x^{99}+x^{100}=1+sum_{n=1}^{100}(x^n+x^{-n})
                  ge201$$
                  with equality if $x=1$.






                  share|cite|improve this answer









                  $endgroup$



                  You can instead minimise the reciprocal of your quantity, viz.,
                  $$frac{1+x+x^2+cdots+x^{200}}{x^{100}}=x^{-100}+x^{-99}+cdots+x^{99}+x^{100}.$$
                  One only needs the two-variable AM/GM inequality to do this, just in the
                  form $y+y^{-1}ge2$ for $y>0$, for
                  $$x^{-100}+x^{-99}+cdots+x^{99}+x^{100}=1+sum_{n=1}^{100}(x^n+x^{-n})
                  ge201$$
                  with equality if $x=1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 30 at 5:43









                  Lord Shark the UnknownLord Shark the Unknown

                  108k1162135




                  108k1162135























                      4












                      $begingroup$

                      For $x>0$, we have from the AM-GM inequality



                      $$begin{align}
                      sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
                      &=201 sqrt[201]{x^{20100}}\\
                      &=201x^{100}
                      end{align}$$



                      Hence, we see that



                      $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Please let me know how I can improve my answer. I really want to give you the best answer I can.
                        $endgroup$
                        – Mark Viola
                        Jan 30 at 5:02
















                      4












                      $begingroup$

                      For $x>0$, we have from the AM-GM inequality



                      $$begin{align}
                      sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
                      &=201 sqrt[201]{x^{20100}}\\
                      &=201x^{100}
                      end{align}$$



                      Hence, we see that



                      $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Please let me know how I can improve my answer. I really want to give you the best answer I can.
                        $endgroup$
                        – Mark Viola
                        Jan 30 at 5:02














                      4












                      4








                      4





                      $begingroup$

                      For $x>0$, we have from the AM-GM inequality



                      $$begin{align}
                      sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
                      &=201 sqrt[201]{x^{20100}}\\
                      &=201x^{100}
                      end{align}$$



                      Hence, we see that



                      $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$






                      share|cite|improve this answer









                      $endgroup$



                      For $x>0$, we have from the AM-GM inequality



                      $$begin{align}
                      sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
                      &=201 sqrt[201]{x^{20100}}\\
                      &=201x^{100}
                      end{align}$$



                      Hence, we see that



                      $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 30 at 4:48









                      Mark ViolaMark Viola

                      134k1278177




                      134k1278177












                      • $begingroup$
                        Please let me know how I can improve my answer. I really want to give you the best answer I can.
                        $endgroup$
                        – Mark Viola
                        Jan 30 at 5:02


















                      • $begingroup$
                        Please let me know how I can improve my answer. I really want to give you the best answer I can.
                        $endgroup$
                        – Mark Viola
                        Jan 30 at 5:02
















                      $begingroup$
                      Please let me know how I can improve my answer. I really want to give you the best answer I can.
                      $endgroup$
                      – Mark Viola
                      Jan 30 at 5:02




                      $begingroup$
                      Please let me know how I can improve my answer. I really want to give you the best answer I can.
                      $endgroup$
                      – Mark Viola
                      Jan 30 at 5:02


















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