What is the maximum value of this fraction?












5












$begingroup$


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?






sequences-and-series algebra-precalculus means geometric-series







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  • 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
    2 hours ago


















5












$begingroup$


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?






sequences-and-series algebra-precalculus means geometric-series







share|cite|improve this question







New contributor




user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    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
    2 hours ago
















5












5








5


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?






sequences-and-series algebra-precalculus means geometric-series







share|cite|improve this question







New contributor




user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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?






sequences-and-series algebra-precalculus means geometric-series




sequences-and-series algebra-precalculus means geometric-series






share|cite|improve this question







New contributor




user69284 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




user69284 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




share|cite|improve this question






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Check out our Code of Conduct.









asked 2 hours ago









user69284user69284

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New contributor




user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





user69284 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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Check out our Code of Conduct.








  • 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
    2 hours ago
















  • 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
    2 hours ago










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
2 hours ago






$begingroup$
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
2 hours ago












3 Answers
3






active

oldest

votes


















5












$begingroup$

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









$endgroup$





















    3












    $begingroup$

    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
      1 hour ago



















    1












    $begingroup$

    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









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      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












        $begingroup$

        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












          5








          5





          $begingroup$

          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 2 hours ago









          jmerryjmerry

          5,987718




          5,987718























              3












              $begingroup$

              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
                1 hour ago
















              3












              $begingroup$

              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
                1 hour ago














              3












              3








              3





              $begingroup$

              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 2 hours ago









              Mark ViolaMark Viola

              131k1275171




              131k1275171












              • $begingroup$
                Please let me know how I can improve my answer. I really want to give you the best answer I can.
                $endgroup$
                – Mark Viola
                1 hour ago


















              • $begingroup$
                Please let me know how I can improve my answer. I really want to give you the best answer I can.
                $endgroup$
                – Mark Viola
                1 hour ago
















              $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can.
              $endgroup$
              – Mark Viola
              1 hour ago




              $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can.
              $endgroup$
              – Mark Viola
              1 hour ago











              1












              $begingroup$

              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









              $endgroup$


















                1












                $begingroup$

                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









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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 1 hour ago









                  Lord Shark the UnknownLord Shark the Unknown

                  103k1160132




                  103k1160132






















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