What is the maximum value of this fraction?












4












$begingroup$


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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  • 1




    $begingroup$
    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
    25 mins ago


















4












$begingroup$


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.







$endgroup$








  • 1




    $begingroup$
    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
    25 mins ago
















4












4








4





$begingroup$


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






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.









asked 45 mins ago









user69284user69284

876




876




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.





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.






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








  • 1




    $begingroup$
    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
    25 mins ago
















  • 1




    $begingroup$
    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
    25 mins ago










1




1




$begingroup$
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
25 mins 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
25 mins ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

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









$endgroup$





















    2












    $begingroup$

    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
      8 mins ago











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

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    3












    $begingroup$

    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












      $begingroup$

      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












        3








        3





        $begingroup$

        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 28 mins ago









        jmerryjmerry

        5,957718




        5,957718























            2












            $begingroup$

            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
              8 mins ago
















            2












            $begingroup$

            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
              8 mins ago














            2












            2








            2





            $begingroup$

            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 22 mins 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
              8 mins 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
              8 mins 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
            8 mins 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
            8 mins ago










            user69284 is a new contributor. Be nice, and check out our Code of Conduct.










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