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
    56 mins 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
    56 mins ago
















5












5








5


1



$begingroup$


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 1 hour ago









user69284user69284

926




926




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.








  • 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
    56 mins 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
    56 mins 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
56 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
56 mins ago












2 Answers
2






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
      39 mins ago











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

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    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 59 mins 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
              39 mins 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
              39 mins 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 53 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
              39 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
              39 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
            39 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
            39 mins ago










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